7 concepts18 worked examples28 exercises3 exam-level7 figures
What are you here for?
13Inductance, Electromagnetic Oscillations, and AC Circuits
A workshop electromagnet runs quietly on a $12\ \mathrm{V}$ battery. You open the switch, and a blue spark jumps the contacts with a crack. Twelve volts cannot push a spark across air; that takes thousands. Everything in this course so far says the current should simply stop, and instead the coil produces, for a millionth of a second, a voltage a thousand times larger than anything in the room. Nothing was added to the circuit. Something was taken away.
By the end of this section you can put a number on that spark: given the coil and the current it was carrying, you can say how much energy was stored in it, how fast the current dies, and how many volts appear across the gap while it does.
In 60 seconds
A coil resists changes in its own current because its own changing flux drives a voltage back through it; that single fact gives the , the stored energy $\tfrac12 LI^{2}$, the exponential rise and fall in an LR circuit, the free oscillation of an LC pair at $1/\sqrt{LC}$, and the whole of the series LRC circuit driven by an alternating source.
a charged capacitor connected to a coil with no resistance worth counting; charge and current trade places twice per cycle and the total energy stays put
R, L and C in one loop across an alternating source; at $\omega_0 = 1/\sqrt{LC}$ the two reactances cancel, $Z$ falls to $R$ and the current peaks
Three most common mistakes
Treating an inductor as something that opposes current. It opposes a change in current: a coil carrying a steady current behaves like a plain wire, and a coil carrying no current at all is the hardest thing in the circuit to get started.
Adding the voltages across R, L and C arithmetically. They peak at different instants, so they add in quadrature; measured separately they can each exceed the source voltage without anything being wrong.
Using the frequency in hertz where the formula wants the angular frequency. Every reactance, every resonance condition and every oscillation formula on this page takes $\omega$ in radians per second, and $\omega = 2\pi f$.
The two midterms carry twenty per cent each, the final twenty five, quizzes ten and homework five, so on the published weights this material can appear in whichever of those falls after week thirteen. The syllabus gives no finer breakdown than that and none is invented here.
How much time do you have?
10 minutes
You leave able to do the two things that open almost every question here: turn a coil into a number of henrys, and turn that number plus a current into stored energy and an induced voltage.
The 60 second card · Formula card · Self-inductance: a coil fighting its own change · Energy stored in a coil, and where it sits · Mistake ledger
45 minutes
You add the three circuits that everything else is built from: the exponential LR switch on and switch off, the free LC oscillation with its energy trading back and forth, and the driven series LRC circuit with its and .
The 60 second card · Self-inductance: a coil fighting its own change · Energy stored in a coil, and where it sits · Switching a coil on and off: the LR circuit · The LC circuit: charge that sloshes · Driving one element at a time: reactance · The series LRC circuit and resonance · Method boxes · Practice C
full read
You add mutual inductance and the reciprocity that makes transformers possible, the and how it compares with the electric one, damping in a real LC circuit, and the interleaved problems that mix this week with the circuits and the magnetism from earlier.
Everything, in order, including the pretest, the ladder and all four practice tiers
By the end of this section
Compute the mutual inductance of a pair of coils from their geometry, and use it in either direction to turn a rate of change of current in one coil into an induced voltage in the other.
Calculate the self-inductance of a coil or a cable from its geometry, and the voltage that appears across it when its current changes at a stated rate.
Determine the energy stored in an inductor two ways, from the current it carries and from the field it holds, and use the magnetic energy density to say where that energy sits.
Analyse an LR circuit at switch on and at switch off, using the time constant to find a current at a given instant, an instant at which a given current is reached, and the voltage produced when the circuit is broken.
Describe a free LC oscillation quantitatively: its angular frequency, the peak current, the way the total energy divides between capacitor and coil at any instant, and what a resistance does to it.
Evaluate the reactance of a coil and of a capacitor at a given frequency, use it to get the current from an alternating source, and state the phase relation and the average power for each element alone.
Solve a driven series LRC circuit for its impedance, current, phase angle, element voltages and average power, and locate and use its resonance.
Syllabus coverage
Inductance, , and AC Circuits
Mutual inductance between two coils and the reciprocity of the two ways of measuring it; self-inductance from the geometry of a solenoid and of a coaxial cable; the energy stored in an inductor and the energy density of a magnetic field; the LR circuit switched on and switched off, its time constant and the large voltage that appears when the circuit is broken; the free LC oscillation, its angular frequency, the exchange of energy between capacitor and coil, and what a resistance does to it; alternating current through a resistor, a coil and a capacitor separately, with their reactances, phase relations and average powers; and the driven series LRC circuit with its impedance, phase angle, and resonance.
The week line is this one phrase and carries no chapter or section number, so no chapter or section number of the textbook is quoted anywhere on this page.
covered
The complete set of field equations
The term that has to be added to the law relating a magnetic field to the current it encircles once fields are allowed to change with time, and the symmetric set of four equations that results.
Deferred to the following week line. Every circuit on this page is treated with circuit quantities alone, and no field equation is modified here.
deferred
Travelling electromagnetic disturbances
What happens when an oscillating circuit radiates, and the speed at which such a disturbance travels.
Deferred to the following week line. The oscillations here are confined to a loop of wire, and no energy is allowed to leave it except as heat in a resistance.
deferred
Transformers and power transmission
Two coils on a shared core, the turns ratio that relates their voltages, and why long distance transmission is done at high voltage.
Not named on the week line and not treated as examinable. It appears here only as the reason mutual inductance is worth a section: the reciprocity result is stated and used, the turns ratio is not.
off_syllabus
Recall first
Faraday's law of induction
A changing magnetic flux through a circuit drives an electromotive force round it, $\mathcal{E} = -N\,d\Phi_B/dt$, where $\Phi_B = \int\vec B\cdot d\vec A$ is the flux through one turn and $N$ is the number of turns.
Every result in this section is this law applied to a flux that the circuit's own current, or a neighbour's, is producing.
Lenz's law
The induced current flows in whichever direction opposes the change in flux that caused it. It opposes the change, not the flux, so a decreasing flux is maintained rather than cancelled.
It is the whole content of the minus signs here, and the only thing standing between you and a coil that would supply free energy.
Magnetic flux
For a uniform field perpendicular to a flat area, $\Phi_B = BA$, in webers. For a coil of $N$ turns the quantity that matters is the $N\Phi_B$.
Inductance is defined as flux linkage divided by current, so the first line of every inductance calculation is a flux.
The field inside a long solenoid
Well inside a long solenoid the field is uniform and equal to $B = \mu_0 nI$, where $n = N/\ell$ is the number of turns per unit length. Outside it is negligible.
It converts the geometry of a coil into a flux, which is the step that produces the standard formula for the self-inductance of a solenoid.
The field around a long straight wire
At a perpendicular distance $r$ from a long straight current the field is $B = \mu_0I/(2\pi r)$, circling the wire by the right hand grip rule.
It is integrated across the gap of a coaxial cable to get that cable's inductance per unit length.
Energy stored in a capacitor
A capacitor holding charge $q$ at voltage $V$ stores $U = q^{2}/(2C) = \tfrac12 CV^{2} = \tfrac12 qV$, and the electric energy density in the field is $u = \tfrac12\varepsilon_0E^{2}$.
The oscillation here is energy moving between a capacitor and a coil, so half of the bookkeeping is this formula, and the magnetic energy density is compared directly with the electric one.
The loop rule
The potential changes round any closed loop of a circuit add to zero. Travelling through a resistor with the current the change is $-IR$; through a source from minus to plus it is $+\mathcal{E}$.
The differential equations for the LR and LC circuits are the loop rule with one new kind of term in them.
The RC time constant and its exponential
A capacitor charging through a resistor approaches its final charge as $1-e^{-t/RC}$ and discharges as $e^{-t/RC}$. After one time constant the change is sixty three per cent complete, after three time constants ninety five per cent.
The LR circuit produces exactly the same two curves with a different time constant, so the algebra of extracting a time from an exponential is already familiar.
Average power in a resistor
A resistor carrying current $I$ dissipates $P = I^{2}R = V^{2}/R$, all of it as heat.
In an alternating circuit this is the only place the average power goes, which is why the final power formula contains $R$ and not $Z$.
Simple harmonic motion
A mass on a spring obeys $m\,d^{2}x/dt^{2} = -kx$, whose solution is $x = A\cos(\omega t+\varphi)$ with $\omega = \sqrt{k/m}$, and whose energy passes back and forth between $\tfrac12kx^{2}$ and $\tfrac12mv^{2}$.
The LC circuit obeys the same equation with different letters, and every question about it can be answered by translating into the mass and spring you already know.
Try it yourself first (3 questions)
1§13.0 — the direction an induced current chooses●●○○○
Before anything new starts, one question that decides whether the rest of this page will make sense. A flat circular loop of wire lies in the plane of the page. A magnetic field points into the page through it, and someone is steadily turning the field up.
Given
a closed circular loop of wire lying flat in the plane of the page
a magnetic field directed into the page through the loop
the strength of that field is increasing steadily with time
the loop itself is not moving and does not change shape
Find
(a) Which way does the induced current run round the loop, and why?
Hint 1/4
The question is not what the applied field is doing. It is what the loop's own induced field has to do to the total flux, and the answer to that fixes the direction of the current.
Hint 2/4
Lenz's law: the induced current flows so that its own magnetic field opposes the change in flux through the loop.
Hint 3/4
The flux into the page is increasing, so the loop's own field inside the loop must point out of the page to fight that increase. The grip rule then converts that field direction into a direction round the loop.
Hint 4/4
A field out of the page inside the loop needs an anticlockwise current, so that is the answer.
Show solution
We name the change in the flux first and let Lenz's law hand us the direction, rather than chasing the force on individual charges: nothing here is moving, so there is no velocity for the force route to use.
Name the change before naming any current
$$\Phi_B \ \text{into the page},\qquad \frac{d\Phi_B}{dt} > 0$$
the loop and its area are fixed, so the only thing changing the flux is the field itself, and it is going up
Apply Lenz's law to get the direction of the induced field
$$\vec B_{\rm induced}\ \text{out of the page inside the loop}$$
the induced field opposes the change, and the change is more flux into the page, so the response points the other way
$$\text{out of the page} \Rightarrow \text{anticlockwise current}$$
the grip rule with the thumb pointing out of the page at the reader curls the fingers anticlockwise
Answer $$\boxed{\ \text{anticlockwise as seen on the page}\ }$$
Check
Independent check by energy: if the current ran clockwise it would add to the flux, which would drive more current, which would add more flux, and the loop would supply energy from nothing. Only the anticlockwise choice is consistent with energy conservation.
Every minus sign in this section is this argument compressed into one symbol, so it is worth being able to run it in five seconds rather than reading it off a formula.
2§13.0 — pulling a time out of an exponential●●○○○
The algebra of this section is the algebra of the charging capacitor, so it is worth checking that it still runs. A $5.00\ \mathrm{\mu F}$ capacitor is charged through a $2.00\ \mathrm{k\Omega}$ resistor from a source of fixed voltage, starting from empty.
Given
$C = 5.00\times10^{-6}\ \mathrm{F}$
$R = 2.00\times10^{3}\ \Omega$
the capacitor starts empty and charges towards a final charge $Q_f$
$q(t) = Q_f\left(1-e^{-t/RC}\right)$
Find
(a) Find the time constant.
(b) Find how long it takes for the charge to reach ninety per cent of its final value.
Hint 1/4
The second part is not a new formula. It is the same formula solved for $t$ instead of for $q$, which means taking a logarithm once.
With $R = 2.00\times10^{3}\ \Omega$ and $C = 5.00\times10^{-6}\ \mathrm{F}$ the time constant is $10.0\ \mathrm{ms}$, and ninety per cent means $q/Q_f = 0.900$, so the bracket is $0.100$.
We isolate the exponential and invert it once with a logarithm, rather than stepping through time constants until the charge passes ninety per cent, because the inversion is exact and costs one line.
Order of magnitude check against the standard landmarks: ninety per cent must lie between two time constants, which gives eighty six per cent, and three, which gives ninety five. It does, at 2.30 of them.
3§13.0 — turns per metre against total turns●●○○○
One number in the solenoid formula is the one this whole section keeps getting wrong, so here it is on its own. A solenoid $20.0\ \mathrm{cm}$ long is wound with $300$ turns and carries $2.00\ \mathrm{A}$.
Given
$N = 300$ turns
$\ell = 0.200\ \mathrm{m}$
$I = 2.00\ \mathrm{A}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
$B = \mu_0 nI$ well inside a long solenoid, with $n$ the turns per unit length
Find
(a) What is the field well inside it?
Hint 1/4
There is only one decision in this question: which of the two turn counts the formula is asking for. Make it explicitly before touching the calculator.
Hint 2/4
$B = \mu_0 nI$ with $n = N/\ell$ in turns per metre, so the formula in terms of the total is $B = \mu_0 NI/\ell$.
Hint 3/4
With $N = 300$ over $\ell = 0.200\ \mathrm{m}$, $n = 1500\ \mathrm{m^{-1}}$, and $I = 2.00\ \mathrm{A}$: $B = (4\pi\times10^{-7})(1500)(2.00)$.
Hint 4/4
$B = 3.77\times10^{-3}\ \mathrm{T}$, that is $3.77\ \mathrm{mT}$, about seventy five times the Earth's field.
Show solution
Turns per metre gets its own line before the field formula is touched; feeding the bare turn count into $B=\mu_0 nI$ is the one slip this formula will never warn you about.
carrying the unit through is what makes the error visible: a bare 300 has no unit and would pass unnoticed into the next line
Substitute
$$B = \mu_0 nI = (4\pi\times10^{-7})(1500)(2.00)$$
the field inside a long solenoid does not depend on the radius or on where inside you stand, so no other geometry is needed
$$B = 3.77\times10^{-3}\ \mathrm{T}$$
three significant figures, matching the data
Answer $$\boxed{\ B = 3.77\times10^{-3}\ \mathrm{T}\ }$$
Check
Plausibility: a few millitesla is around a hundred times the Earth's field and a tenth of a fridge magnet, which is the right region for a small air cored coil carrying a couple of amps.
The same $n$ against $N$ decision reappears in the self-inductance formula, where getting it wrong costs a factor of the length rather than a factor of the length once removed.
Notation
symbol
reads as
means
watch out
$L$
L
the self-inductance of a coil, in henrys, defined as the flux linkage per unit of its own current
the same letter is used in earlier sections for a length, and in this section a length is always written $\ell$ for exactly that reason
$M$
M
the mutual inductance of a pair of coils, in henrys, defined as the flux linkage in one per unit current in the other
it belongs to the pair, not to either coil, and it is the same number whichever coil is called the driver
$\mathcal{E}$
script E
an induced electromotive force in volts, which is a voltage and not a force despite the name
for an inductor with no resistance the terminal voltage and the induced electromotive force are the same size, so problems switch freely between calling it $\mathcal{E}$ and calling it $V_L$
$\tau$
tau
the time constant of an LR circuit, $L/R$, in seconds
the earlier circuit work used $\tau = RC$ where the resistance multiplied; here it divides, so a larger resistance makes an LR circuit faster and an RC circuit slower
$\omega_0$
omega nought
the of an LC pair, $1/\sqrt{LC}$, in radians per second
it is a property of the two components alone; when the same pair is driven by a source, $\omega_0$ is the frequency at which resonance happens and $\omega$ is whatever the source is doing
$X_L,\ X_C$
X L and X C
the reactance of a coil and of a capacitor, both in ohms, each relating the voltage across that element to the current through it
an ohm here does not mean a resistance: no average power is consumed in either, and the two respond to frequency in opposite directions
$Z$
Z
the impedance of a circuit in ohms, the single number that relates the source voltage to the current
it is never the sum of the individual ohms; the reactive parts subtract from each other before anything is squared
$\phi$
phi
the phase angle by which the source voltage leads the current, positive for an inductive circuit and negative for a capacitive one
$\cos\phi$ is the power factor and is never negative for a passive circuit, so a sign that survives into a power calculation is a sign that has gone wrong
$q,\ Q_0$
q and Q nought
the instantaneous charge on the capacitor and its peak value during an oscillation
the letter $Q$ is also the standard symbol for the quality factor of a resonant circuit; here the charge always carries a subscript and the quality factor is written out in words wherever it appears
$u$
little u
energy per unit volume stored in a field, in joules per cubic metre
the magnetic form has $\mu_0$ dividing and the electric form has $\varepsilon_0$ multiplying, which is the usual pattern for those two constants and not a slip
$n$
little n
turns per unit length of a solenoid, $N/\ell$, in turns per metre
the self-inductance formula can be written with either $N$ or $n$ and the two differ by a factor of the length; deciding which one is in front of you before substituting is the single most valuable habit in this section
Conventions used here
What the minus sign in the inductance formulas is doing
Every induced voltage on this page is written with a minus sign in the general statement and then used as a magnitude in the arithmetic. The minus sign carries no numerical information: it is Lenz's law in symbols, and it says that the induced voltage opposes the change that produced it. In every worked example the size comes from the formula and the direction comes from one sentence of Lenz's law reasoning, never from the sign of a number that has been carried through three lines of algebra.
Which instant a symbol refers to in a circuit that is changing
A subscript zero means a peak value, so $I_0$ and $V_0$ are the largest values reached in a cycle. A subscript rms means the , which for a sine wave is the peak divided by $\sqrt{2}$. A symbol written as a function of time, $I(t)$ or $q(t)$, is the instantaneous value at that one moment. An ammeter or a voltmeter on an alternating circuit reads root mean square, and unless a problem says otherwise, so does a quoted mains voltage.
How an angular frequency and a frequency are kept apart
The angular frequency $\omega$ is in radians per second and the frequency $f$ is in hertz, with $\omega = 2\pi f$ always. Every reactance, every resonance condition and every oscillation formula on this page is written in $\omega$, and every laboratory instrument and every mains supply is quoted in $f$. The conversion is done once, at the start of a problem, and written on its own line so that the factor of $2\pi$ cannot go missing.
Where the phase angle is measured from
The phase angle $\phi$ of a series circuit is the angle by which the source voltage leads the current. A positive $\phi$ therefore means the circuit is inductive and the current lags behind the voltage; a negative $\phi$ means it is capacitive and the current leads. The current is used as the reference in every series circuit here because it is the one quantity that is the same everywhere in the loop.
What an ideal coil, capacitor and source mean here
An inductor has no resistance unless the problem gives it one, in which case that resistance is drawn as a separate resistor in series with a pure inductance. A capacitor leaks no charge, a source has no internal resistance unless stated, and connecting wires have neither resistance nor inductance. A solenoid is long enough for the uniform field formula to hold, and the flux through every one of its turns is the same.
Which constants and how many digits this section keeps
The permeability of free space is $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$, and where the combination appears it is used as $\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$. The permittivity of free space is $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$. Every answer is quoted to three significant figures, which is what the data given in these problems supports, and no answer is carried to more.
How a current direction is fixed in an oscillating circuit
A positive current is one that increases the charge on the upper plate of the capacitor as drawn, so that $I = dq/dt$ with no sign to remember. In an oscillation the current reverses twice a cycle, and a negative value of $I$ then means the charge is flowing the other way round the loop. No physical answer on this page depends on which plate was called the upper one, and any quantity that would change with that choice is quoted as a magnitude.
13.1Mutual inductance: one coil talking to another
A single number fixed by geometry converts a rate of change of current in one coil into a voltage in the other.
The spark came out of a coil, so the missing quantity links a current to the flux it makes, and two coils show it most plainly.
Solvable with what we have
Turn a rate of change of flux into a voltage.
Turn a field and an area into a flux.
Produce the field well inside a long solenoid.
Not solvable yet
Say what voltage a neighbouring coil gets when this current changes at $25.0\ \mathrm{A/s}$.
Say how many volts appear when a coil's own current stops.
Say why a coil and a capacitor ring at one frequency.
Faraday's law written straight down for the second coil gives its voltage in terms of the flux through it. Correct and useless: what the problem hands you is a rate of change of current somewhere else.
Why it fails
Faraday's law speaks about flux; every practical question here is asked about currents. Between them sits a missing conversion factor, and it can depend only on the shapes and positions of the two coils, since nothing else stays fixed while the current changes.
DefinitionDefinition 13.1: mutual inductance
Conditions
the two coils are rigid and fixed in place relative to each other
there is no material nearby whose magnetic response changes with the current, so a doubled current means a doubled flux
the flux $\Phi_{21}$ is the flux through one turn of coil 2 produced by the current in coil 1, and it is multiplied by $N_2$ to give the flux linkage
Count the flux one coil sends through every turn of the other, divide by the current that made it, and you have a number in henrys belonging to the pair. Multiply it by how fast either current changes, in amps per second, to get the voltage in the other coil. The last equality is the surprise: one number serves both directions.
Proof
Start from the definition. Because the flux is proportional to the current that makes it, the ratio $N_2\Phi_{21}/I_1$ does not depend on how big the current happens to be, and it is that constancy which makes the definition worth writing down.
Rearranged, the definition says $N_2\Phi_{21} = MI_1$: the flux linkage in coil 2 is the mutual inductance times the current in coil 1.
Now differentiate with respect to time. The geometry is fixed, so $M$ is a constant and comes out of the derivative: $N_2\,d\Phi_{21}/dt = M\,dI_1/dt$.
Faraday's law for coil 2 reads $\mathcal{E}_2 = -N_2\,d\Phi_{21}/dt$, and the left side of the previous line is exactly that quantity with the sign removed. Substituting gives $\mathcal{E}_2 = -M\,dI_1/dt$.
Notice what has happened: a law about flux has been turned into a law about current, at the price of one number that has to be worked out once from the geometry. That trade is made three more times in this section.
The reciprocity $M_{12} = M_{21}$ is quoted here rather than derived, because a general derivation needs the energy of the pair of circuits and would take a page. It is worth trusting for a practical reason as well as a theoretical one: the awkward direction is usually the impossible calculation, and reciprocity lets you compute the easy one instead.
The chain the definition compresses into one letter. A changing $\textcolor{#d1690a}{\text{current in coil 1}}$ makes a changing $\textcolor{#1f6feb}{\text{flux}}$, and that flux, threading every turn of $\textcolor{#8250df}{\text{coil 2}}$, drives a voltage there. Mutual inductance is the fixed number that lets you jump from the first link straight to the last.
Looks like this, but is not
Two big coils side by side must have a large mutual inductance. They are close, they are wide, and one is making a strong field right where the other sits.
Turn the second coil through a right angle, so its plane contains the first one's axis, and the mutual inductance drops to zero. The field now runs across its face instead of through it: every line entering one half leaves through the other, and the net linkage is nothing. Size and closeness are not enough; what counts is flux that threads the turns.
turns on the small coil, $N_2$
mutual inductance
induced voltage in coil 2
$25$
$2.51\times10^{-5}\ \mathrm{H}$
$0.628\ \mathrm{mV}$
$50$
$5.03\times10^{-5}\ \mathrm{H}$
$1.26\ \mathrm{mV}$
$100$
$1.01\times10^{-4}\ \mathrm{H}$
$2.51\ \mathrm{mV}$
$200$
$2.01\times10^{-4}\ \mathrm{H}$
$5.03\ \mathrm{mV}$
Doubling the secondary doubles both the mutual inductance and the voltage, with nothing else in the geometry touched: $M$ is straight proportional to $N_2$, and to $N_1$ for the same reason. The length sits underneath, so stretching the same $1000$ turns over a whole metre would halve every entry, because it is turns per metre that make the field. Notice what is still missing from the table: the size of the small coil never appears, and neither does any angle. Tilting a winding that goes round the cylinder changes nothing, because the slanted loop's own area grows by exactly the factor by which the perpendicular component of the field falls, and the two cancel. That is a feature of a winding that encircles the solenoid; two free standing coils facing each other, as in the counterexample above, are a different arrangement and do lose their linkage when one is turned.
A small coil wound round the middle of a solenoid
A long solenoid is $0.500\ \mathrm{m}$ long, has $1000$ turns and a cross sectional area of $4.00\ \mathrm{cm^{2}}$. A second coil of $50$ turns is wound tightly around its middle. Find the mutual inductance of the pair, and the voltage induced in the small coil when the solenoid's current is changing at $25.0\ \mathrm{A/s}$.
small coil: $N_2 = 50$ turns, wound tightly round the middle
$dI_1/dt = 25.0\ \mathrm{A/s}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
the mutual inductance, and the induced voltage in the small coil
Solution
We drive the solenoid and read the small coil, not the other way round: the reciprocal arrangement would need the field of a fifty turn coil at every point inside the solenoid, which is an integral nobody wants to do, while the solenoid's own field is a one line formula.
Drive the pair from the easy end
$$M = \frac{N_2\Phi_{21}}{I_1}$$
the definition can be evaluated by driving either coil, and only one of the two directions is workable here: the field inside the solenoid is a formula, the field of a fifty turn coil at every point inside the solenoid is not
the field outside the solenoid is negligible, so the area that counts is the solenoid's cross section and not the area of the small coil, even though it is the small coil's flux being counted
Independent check by units on the symbolic result: $\mu_0$ carries $\mathrm{T\,m/A}$, the area carries $\mathrm{m^{2}}$ and the length divides one metre out, leaving $\mathrm{T\,m^{2}/A}$, which is a weber per amp, which is a henry. The turn counts are pure numbers and cannot spoil it.
Three lines of geometry and one substitution. The whole difficulty was choosing which coil to call the driver.
Note what the answer does not contain: the area of the small coil, its length, or the wire it is made of. Only the solenoid's cross section appears, because that is the only region where there is any field to catch.
Using reciprocity to get the measurement you cannot make
Two coils are fixed on a bench. When the current in coil 1 is changed at $45.0\ \mathrm{A/s}$, a voltmeter across coil 2 reads $12.0\ \mathrm{mV}$. Coil 1 is then left open and the current in coil 2 is changed at $30.0\ \mathrm{A/s}$ instead. What does a voltmeter across coil 1 read?
second measurement: $dI_2/dt = 30.0\ \mathrm{A/s}$
the coils are not moved between the two measurements
Find
the voltage induced in coil 1 in the second arrangement
Solution
No geometry is given, so we run the definition backwards from the first measurement to get the single number $M$ and then push it through reciprocity; modelling the two coils' shapes would need dimensions the problem never supplies.
no geometry is given and none is needed; the definition can be run backwards from a measurement, which is how mutual inductances are found for real coils whose shapes nobody wants to integrate over
Independent check by proportion rather than by repeating the arithmetic: the second rate of change is two thirds of the first, and $8.00$ is two thirds of $12.0$, exactly as a single shared constant of proportionality demands.
One division and one multiplication, once the reciprocity result is believed.
This is the pattern to carry forward: when a problem gives you a measurement rather than a geometry, the definition is a way of extracting the constant, and the constant then answers questions the original measurement was never set up to answer.
Checkpoint
§13.1 — mutual inductance from a switched off current●●○○○
Thirty seconds, one substitution. Two coils on a shared iron free frame have a mutual inductance of $0.240\ \mathrm{H}$. The current in the first coil is switched off, falling steadily from $4.00\ \mathrm{A}$ to zero in $20.0\ \mathrm{ms}$.
Given
$M = 0.240\ \mathrm{H}$
the current in coil 1 falls steadily from $4.00\ \mathrm{A}$ to $0$
the fall takes $20.0\times10^{-3}\ \mathrm{s}$
Find
(a) What is the size of the voltage induced in the second coil while this happens?
Hint 1/4
Steadily means the rate of change is one number for the whole interval, so no calculus is needed: a difference divided by a time will do.
Hint 2/4
$|\mathcal{E}_2| = M\,|dI_1/dt|$, and for a steady change $|dI_1/dt| = |\Delta I_1|/\Delta t$.
Hint 3/4
With $\Delta I_1 = 4.00\ \mathrm{A}$ over $\Delta t = 20.0\ \mathrm{ms}$ the rate is $200\ \mathrm{A/s}$, and $M = 0.240\ \mathrm{H}$.
The rate of change is worked out on its own line before the voltage formula is opened, because this question is lost in the millisecond and not in the multiplication.
Independent check by scaling: switching the same current off in a tenth of the time would give ten times the voltage, which is the behaviour a spark at the contacts is evidence for, and there is nothing in the formula to stop it.
⚠ Treating the mutual inductance as belonging to one coil
the symbol is written next to whichever coil is being driven, so it starts to look like a property of that coil in the way resistance is a property of a resistor
wrong$$M_{12} \neq M_{21}$$
right$$M_{12} = M_{21} = M$$
⚠ Using the area of the wrong coil
the flux being counted belongs to coil 2, so its own area feels like the right one to reach for
wrong$$\Phi_{21} = B_1A_2$$
right$$\Phi_{21} = B_1A_{\rm solenoid}$$
⚠ Multiplying by the current instead of by its rate of change
the definition contains a current and the formula for the voltage does not, and the two get merged into one half remembered line
wrong$$\mathcal{E}_2 = -MI_1$$
right$$\mathcal{E}_2 = -M\dfrac{dI_1}{dt}$$
13.2Self-inductance: a coil fighting its own change
A coil's own changing current changes its own flux, and the voltage that produces always opposes whatever is being attempted.
Nothing in the last argument required two coils. Let the flux a coil makes thread its own turns and the same conversion factor appears, with a new consequence: the coil now resists the change it is itself producing.
DefinitionDefinition 13.2: self-inductance, and the voltage across a coil
Conditions
the coil is rigid and no magnetic material near it changes its response with current, so that flux stays proportional to current
$\Phi_B$ is the flux through a single turn and $N\Phi_B$ is the flux linkage of the whole coil
the solenoid form assumes a long solenoid of uniform cross section $A$ and length $\ell$, with the same flux through every turn
the coil has no resistance of its own; a real coil is drawn as this ideal inductance in series with a separate resistor
Take the flux through one turn, multiply by the number of turns, divide by the current making it, and the answer is the self-inductance in henrys. One henry means that changing the current by one amp per second produces one volt across the coil. The minus sign says the voltage opposes the change, and the solenoid formula says the turns count twice: once in making the field and once in catching it.
Proof
The solenoid formula is worth deriving once, because it is the only inductance most problems ever want and because the squared turn count surprises people.
The field well inside a long solenoid carrying current $I$ is $B = \mu_0nI = \mu_0NI/\ell$, uniform and along the axis.
The flux through one turn is $\Phi_B = BA = \mu_0NIA/\ell$, since the field is uniform over the cross section and perpendicular to it.
The flux linkage is $N$ times that, because the same flux passes through every one of the turns: $N\Phi_B = \mu_0N^{2}IA/\ell$.
Divide by $I$, which cancels: $L = \mu_0N^{2}A/\ell$. The turn count entered twice for two different reasons, which is why it is squared, and the current has vanished, which is why $L$ is a property of the coil and not of what it is doing.
Written with the turns per metre instead, $N = n\ell$ gives $L = \mu_0n^{2}A\ell = \mu_0n^{2}V$, where $V$ is the volume the field occupies. That form is the one to keep in mind when the energy appears two concepts from now.
The one sentence worth memorising, drawn twice. The $\textcolor{#8250df}{\text{induced voltage}}$ is set by what the $\textcolor{#d1690a}{\text{current}}$ is doing, not by which way it flows: it fights a rise and it props up a fall.
Looks like this, but is not
An inductor is a component that opposes current, the way a resistor does. It has ohms in its formulas, it limits the current when you switch on, and it makes circuits sluggish.
Put a steady current through an ideal inductor and it produces no voltage at all: with $dI/dt = 0$ the formula gives zero, and the coil is indistinguishable from a piece of wire. A resistor with $5\ \mathrm{A}$ through it always drops a voltage; an inductor with $5\ \mathrm{A}$ through it drops nothing unless that $5$ is on its way somewhere. What an inductor opposes is the verb, not the noun.
turns
turns per metre
self-inductance
$200$
$800\ \mathrm{m^{-1}}$
$0.142\ \mathrm{mH}$
$400$
$1600\ \mathrm{m^{-1}}$
$0.569\ \mathrm{mH}$
$800$
$3200\ \mathrm{m^{-1}}$
$2.27\ \mathrm{mH}$
$1600$
$6400\ \mathrm{m^{-1}}$
$9.10\ \mathrm{mH}$
Every doubling of the winding multiplies the inductance by four, not by two. The extra turns make a stronger field and then catch more of it, and the two effects are the same effect counted twice.
An air cored solenoid, and the volts it makes when you switch it off
A solenoid $25.0\ \mathrm{cm}$ long is wound with $400$ turns on a form of radius $1.50\ \mathrm{cm}$, with air inside. Find its self-inductance. It is carrying $3.00\ \mathrm{A}$ when a switch is opened and the current falls steadily to zero in $8.00\ \mathrm{ms}$; find the voltage across the coil while that happens.
$I$ falls steadily from $3.00\ \mathrm{A}$ to $0$ in $8.00\times10^{-3}\ \mathrm{s}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
the self-inductance, and the voltage across the coil during the switch off
Solution
We use the $N$ form of the solenoid result because the problem states a total turn count; going through turns per metre would also work but adds a line and one more chance to divide by the wrong length.
Turn the winding into a cross section and a turn count
the $N$ form is chosen because the problem states a total turn count; had it stated turns per centimetre, the $n$ form would have been the honest one and the length would have entered differently
Independent check of $L$ through the volume form: $n = 400/0.250 = 1600\ \mathrm{m^{-1}}$ and the volume is $A\ell = 1.77\times10^{-4}\ \mathrm{m^{3}}$, so $\mu_0n^{2}A\ell = (4\pi\times10^{-7})(1600)^{2}(1.77\times10^{-4}) = 5.69\times10^{-4}\ \mathrm{H}$, reached by a different route with different intermediate numbers.
One area, one substitution, one ratio. Nothing here needed calculus, because the current was said to fall steadily.
Two hundred millivolts is unremarkable, and that is the point worth holding on to: this switch off took eight milliseconds. The hook's spark is the same calculation with a switch that opens in a microsecond, and the voltage scales in exact proportion.
The inductance of a coaxial cable, from an integral over the gap
A coaxial cable has an inner conductor of radius $1.00\ \mathrm{mm}$ and an outer sheath of inner radius $4.00\ \mathrm{mm}$, with the current running out along the core and back along the sheath. Find its self-inductance per unit length, and the inductance of a $20.0\ \mathrm{m}$ length of it.
$b = 4.00\times10^{-3}\ \mathrm{m}$ (inner surface of the sheath)
equal and opposite currents $I$ in core and sheath
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
the field between the conductors is $B = \mu_0I/(2\pi r)$, and outside the sheath it is zero
Find
the self-inductance per metre, and for $20.0\ \mathrm{m}$
Solution
There is no ready made formula for a coaxial line, so we go back to flux and integrate across the gap; picking a mean radius and multiplying would be quicker and wrong, because the field varies as $1/r$ over the whole gap.
Choose the surface the flux passes through
$$dA = \ell\,dr,\qquad a\le r\le b$$
the field circles the axis, so the surface that catches it is the flat strip lying between the two conductors along the length of the cable; a disc across the cable would catch none of it
Integrate, because the field is not uniform over that strip
Limiting case check rather than a repeat of the arithmetic: as the sheath is brought down towards the core, $b/a\to1$, the logarithm goes to zero and so does the inductance, which is right, because there is then no space left for any flux to sit in.
One integral, and it is the only one in this section. Every other inductance here comes from a uniform field.
Notice that the answer depends on the radii only through their ratio. Doubling both leaves the inductance untouched, which is a useful sanity check on any cable formula and an easy exam question to disguise.
Checkpoint
§13.2 — voltage across a coil from a rate of change●●○○○
Thirty seconds. A coil of self-inductance $45.0\ \mathrm{mH}$ has its current raised steadily from $2.00\ \mathrm{A}$ to $5.00\ \mathrm{A}$ over $60.0\ \mathrm{ms}$.
Given
$L = 45.0\times10^{-3}\ \mathrm{H}$
the current rises steadily from $2.00\ \mathrm{A}$ to $5.00\ \mathrm{A}$
the rise takes $60.0\times10^{-3}\ \mathrm{s}$
Find
(a) What is the size of the voltage across the coil while the current is rising?
Hint 1/4
Only the change matters. The fact that the current is nowhere near zero is a distraction that the formula ignores.
Hint 2/4
$|\mathcal{E}| = L\,|dI/dt|$, with the rate taken as the change in current over the time it takes.
Hint 3/4
The change is $5.00-2.00 = 3.00\ \mathrm{A}$ over $60.0\ \mathrm{ms}$, which is $50.0\ \mathrm{A/s}$, and $L = 45.0\ \mathrm{mH}$.
Independent check by scaling from the definition of the henry: one henry at one amp per second gives one volt, so $0.045$ of a henry at $50$ amps per second gives $0.045\times50$ volts directly.
⚠ Putting the total turns where turns per metre belong
the two forms of the solenoid formula look almost identical and both are correct, so the wrong one gets copied from memory under time pressure
the resistor habit is strong: there, the voltage is proportional to the current itself
wrong$$V_L = LI$$
right$$V_L = L\left|\dfrac{dI}{dt}\right|$$
13.3Energy stored in a coil, and where it sits
Building a current against a coil's own opposition costs work, and that work waits in the magnetic field until the current dies.
If a coil fights every attempt to change its current, then getting a current going through one costs work, and the question is where that work goes and how to get it back.
TheoremResult 13.3: the energy in an inductor and in a magnetic field
Conditions
the inductance is constant, which fails for an iron cored coil driven towards saturation
the coil has no resistance of its own, so that none of the work done is lost as heat on the way in
the energy density form assumes the field is uniform over the volume $V$; where it is not, the volume has to be cut into pieces and the contributions added
$U$ is the energy that comes back out when the current is brought to zero, not the energy dissipated in getting it there
$$\boxed{\ U = \tfrac{1}{2}LI^{2},\qquad u = \frac{U}{V} = \frac{B^{2}}{2\mu_0}\ }$$
Half the inductance times the square of the current is the energy sitting in a coil, in joules. The second form says the same number differently: square the field, divide by twice the permeability of free space, and you have the joules in every cubic metre the field occupies. Doubling the current stores four times the energy, and the square is the reason a coil is a serious thing to switch off.
Proof
Work out what it costs to build the current up from nothing. At the instant the current is $I$ and rising at $dI/dt$, the coil is pushing back with $\mathcal{E} = L\,dI/dt$, so the source has to supply power $P = \mathcal{E}I = LI\,dI/dt$ just to keep going.
The energy delivered in a short time is $dU = P\,dt = LI\,dI$. This is the key line: the awkward $dt$ cancels and what is left is an integral over current, which is a quantity we control.
Integrating from zero current to the final value $I$: $U = \int_0^{I}LI'\,dI' = \tfrac12LI^{2}$. The factor of a half comes from the integration, exactly as it does for a capacitor and for a spring, and for the same reason: the opposition grows as you push.
Now ask where the energy is. Take a long solenoid, where $L = \mu_0n^{2}A\ell$ and $B = \mu_0nI$, so that $I = B/(\mu_0n)$.
Substitute both: $U = \tfrac12(\mu_0n^{2}A\ell)\dfrac{B^{2}}{\mu_0^{2}n^{2}} = \dfrac{B^{2}}{2\mu_0}A\ell$. Every reference to the winding has cancelled and only the field and the volume it occupies are left.
Since $A\ell$ is the volume inside the solenoid, the energy per unit volume is $u = B^{2}/(2\mu_0)$. It was derived from a solenoid but it contains nothing about solenoids, which is the usual sign that a result is more general than its derivation.
Stored energy against current for the $0.569\ \mathrm{mH}$ coil of the previous concept. The $\textcolor{#1f6feb}{\text{curve}}$ is a parabola, so the two $\textcolor{#8250df}{\text{marked points}}$, one at twice the current of the other, differ by a factor of four and not of two.
Looks like this, but is not
The energy must be in the moving charges, since it vanishes the moment the current stops. A current is charges in motion, motion is kinetic energy, and the timing fits perfectly.
The drift speed of electrons in a wire is under a millimetre per second, and their total kinetic energy at that speed is smaller than the stored energy by many powers of ten. The energy is in the field, and the strongest evidence is that two completely independent calculations, one using only the coil's inductance and current, the other using only the field strength and the volume it fills, give the same number every time.
field
energy density
comparable with
$5.0\times10^{-5}\ \mathrm{T}$, the Earth
$9.95\times10^{-4}\ \mathrm{J/m^{3}}$
a gram falling ten centimetres, spread over a whole cubic metre
$0.0100\ \mathrm{T}$, a fridge magnet
$39.8\ \mathrm{J/m^{3}}$
air at its breakdown field, which stores $39.8\ \mathrm{J/m^{3}}$ electrically
$1.00\ \mathrm{T}$, a laboratory magnet
$3.98\times10^{5}\ \mathrm{J/m^{3}}$
a family car at ninety kilometres an hour, in every cubic metre
$8.00\ \mathrm{T}$, a superconducting magnet
$2.55\times10^{7}\ \mathrm{J/m^{3}}$
about six kilograms of chemical explosive per cubic metre
The density goes as the square of the field, so the last row is not eight times the third but sixty four times it. That squaring is why large magnets are stored behind interlocks and why nobody worries about the Earth's field.
The same energy counted twice: from the coil, and from the field
The solenoid of the previous concept, $25.0\ \mathrm{cm}$ long with $400$ turns and a cross section of $7.07\ \mathrm{cm^{2}}$, has $L = 5.69\times10^{-4}\ \mathrm{H}$ and carries $3.00\ \mathrm{A}$. Find the stored energy from the inductance, then find the field inside, the magnetic energy density, and the total energy in that volume. Check that the two answers agree.
the stored energy by both routes, and whether they agree
Solution
Both routes are run on purpose. The circuit route is the short one and is the answer; the field route is here to show that it is the same energy seen from the other side, not to be a slower way of getting the number.
squaring a small number and dividing by a small number gives something of ordinary size, which is why an order of magnitude check is worth doing here rather than trusting the powers of ten
the same number as the circuit route, reached without ever mentioning the inductance
Answer $$\boxed{\ U = 2.56\ \mathrm{mJ}\ \text{by both routes};\quad B = 6.03\ \mathrm{mT},\ u = 14.5\ \mathrm{J/m^{3}}\ }$$
Check
The agreement of the two routes is itself the verification, and it is a genuine one: the first used $L$ and $I$ and nothing else, the second used $N$, $\ell$, $A$ and $I$ and never mentioned $L$. Two different sets of numbers arriving at $2.56\ \mathrm{mJ}$ is not a coincidence that survives an arithmetic slip.
Two routes to one answer, four substitutions in total. In an exam you would do one and use the other only if a check were free.
This is the pattern for every stored energy question in this section: if the geometry is given, both routes are open, and the one you do not need is the one that checks the one you do.
Why energy is stored magnetically rather than electrically
Compare the energy density of a $1.00\ \mathrm{T}$ magnetic field with that of the strongest electric field dry air can hold, $3.0\times10^{6}\ \mathrm{V/m}$, above which it breaks down and sparks.
Given
$B = 1.00\ \mathrm{T}$
$E = 3.0\times10^{6}\ \mathrm{V/m}$, the breakdown field of air
$u_B = B^{2}/(2\mu_0)$ and $u_E = \tfrac12\varepsilon_0E^{2}$
the two energy densities and the ratio between them
Solution
We compare energy densities rather than energies, because the two cases share no volume to compare over, and we hand the electric side the strongest field air can hold so that the verdict cannot be blamed on an unfair choice.
the breakdown field is deliberately chosen because it is the most favourable case: any stronger and the air conducts and the stored energy leaves as a spark
Plausibility of the electric figure by an independent route: a cubic metre of air at breakdown holding forty joules is about the energy of a two kilogram book falling two metres, which matches the modest crack of a static spark rather than an explosion.
Two substitutions and one division, with the whole difficulty in choosing a fair field for the electric case.
This is why energy in electrical engineering is stored in coils and in batteries rather than in capacitors, and it is worth remembering as a rough fact rather than as a formula: a laboratory magnetic field beats the best air can do electrically by about ten thousand.
Checkpoint
§13.3 — stored energy from inductance and current●○○○○
Thirty seconds. A coil of self-inductance $0.150\ \mathrm{H}$ is carrying a steady current of $4.00\ \mathrm{A}$.
Given
$L = 0.150\ \mathrm{H}$
$I = 4.00\ \mathrm{A}$, steady
the coil's own resistance is negligible
Find
(a) How much energy is stored in it?
(b) By what factor would that change if the current were raised to $8.00\ \mathrm{A}$?
Hint 1/4
The word steady is there to reassure you, not to complicate things: a steady current means no induced voltage, but the stored energy does not care.
Hint 2/4
$U = \tfrac12LI^{2}$, and since the current is squared, a factor on the current becomes that factor squared on the energy.
Hint 3/4
With $L = 0.150\ \mathrm{H}$ and $I = 4.00\ \mathrm{A}$: $U = \tfrac12(0.150)(16.0)$, and doubling the current means a factor of $2^{2}$.
Hint 4/4
$U = 1.20\ \mathrm{J}$, and doubling the current makes it four times larger, $4.80\ \mathrm{J}$.
Show solution
The second part is done as a ratio rather than a second substitution: the inductance cancels, so the factor of four would stand even if $L$ had never been given.
the inductance is the same in both cases, so it cancels in the ratio and there is no need to substitute a second time
Answer $$\boxed{\ U = 1.20\ \mathrm{J},\qquad \text{factor } 4\ }$$
Check
Independent check of the second part by direct substitution: $\tfrac12(0.150)(8.00)^{2} = 4.80\ \mathrm{J}$, which is indeed four times $1.20\ \mathrm{J}$.
⚠ Dropping the factor of a half
the half comes from an integration that is done once and then never seen again, so there is nothing in the final formula to remind you of it
wrong$$U = LI^{2}$$
right$$U = \tfrac12LI^{2}$$
⚠ Multiplying the energy density by the wrong volume
the coil has an obvious outside volume, and the field does not live in it
wrong$$U = \frac{B^{2}}{2\mu_0}\times(\text{volume of the whole coil former})$$
right$$U = \frac{B^{2}}{2\mu_0}\times(\text{volume where the field actually is}) = \frac{B^{2}}{2\mu_0}A\ell$$
⚠ Putting $\mu_0$ in the numerator of the energy density
the electric form has $\varepsilon_0$ multiplying, and the two are assumed to behave alike
wrong$$u = \tfrac12\mu_0B^{2}$$
right$$u = \frac{B^{2}}{2\mu_0}$$
13.4Switching a coil on and off: the LR circuit
A coil and a resistor make the current climb and fall exponentially, on a clock set by L divided by R.
So far the current has been changing at a rate somebody handed us. Put the coil in a real circuit with a resistor and a battery, and the circuit decides that rate for itself.
TheoremResult 13.4: current in an LR circuit, and its time constant
Conditions
a single loop containing a source of fixed voltage $V_0$, a resistance $R$ and an inductance $L$ in series
the growth form assumes the current starts at zero; the decay form assumes the source has been removed and the loop closed on itself
$R$ is the total resistance of the loop, which includes the coil's own winding resistance when the problem gives one
the switch is ideal, so the geometry of the circuit does not change while the current is changing
Close the switch and the current does not jump; it climbs towards the value the resistor alone would allow, getting sixty three per cent of the way there in one time constant. Open it and the current does not stop; it falls away with the same clock. That clock is the inductance divided by the resistance, so a bigger coil is slower and, against every instinct from the capacitor work, a bigger resistor is faster.
Proof
Apply the loop rule going round with the current. Through the source it gains $V_0$, through the resistor it loses $IR$, and across the inductor it loses $L\,dI/dt$, since the coil opposes the rise: $V_0 - IR - L\,dI/dt = 0$.
Read the two ends off before solving anything. At the instant of closing, $I = 0$, so $L\,dI/dt = V_0$: the whole source voltage is across the coil and the current is rising as fast as it ever will. After a long time nothing is changing, so $dI/dt = 0$ and $I = V_0/R$: the coil has become a piece of wire.
Between those two ends, rearrange and separate: $\dfrac{dI}{V_0/R - I} = \dfrac{R}{L}\,dt$.
Integrating from $I = 0$ at $t = 0$ gives $-\ln\!\left(1 - \dfrac{IR}{V_0}\right) = \dfrac{R}{L}t$, and exponentiating gives the growth form with $\tau = L/R$.
For the decay there is no source: $-IR - L\,dI/dt = 0$, which is $dI/I = -(R/L)\,dt$ and integrates immediately to $I = I_0e^{-t/\tau}$, with the same time constant.
Check that $L/R$ is a time. A henry is a volt second per amp and an ohm is a volt per amp, so their ratio is a second. This is worth doing once, because $L/R$ does not look like a time and $RC$ at least contains a hint of one.
The switch on curve, with the meaning of the time constant built in. The $\textcolor{#1f6feb}{\text{current}}$ climbs towards $V/R$ but never quite arrives; the $\textcolor{#8250df}{\text{tangent at the start}}$ would arrive at exactly one time constant, which is what the time constant is.
Looks like this, but is not
Closing the switch puts the source across the resistor, so the current is $V_0/R$ straight away. The coil has no resistance, so it should not affect anything.
A current appearing instantly would mean an infinite rate of change, and the coil would then produce an infinite voltage to fight it. At the first instant the current is exactly zero and the entire source voltage sits across the coil, which is the opposite of the guess: at that moment the coil behaves like a break in the circuit, not like a wire. It becomes a wire only after several time constants, once nothing is changing any more.
time
current
voltage across the coil
$0$
$0$
$12.0\ \mathrm{V}$
$\tau = 6.00\ \mathrm{ms}$
$0.253\ \mathrm{A}$
$4.41\ \mathrm{V}$
$2\tau = 12.0\ \mathrm{ms}$
$0.346\ \mathrm{A}$
$1.62\ \mathrm{V}$
$3\tau = 18.0\ \mathrm{ms}$
$0.380\ \mathrm{A}$
$0.597\ \mathrm{V}$
$5\tau = 30.0\ \mathrm{ms}$
$0.397\ \mathrm{A}$
$0.0809\ \mathrm{V}$
The two columns are mirror images, and they have to be: whatever voltage is not across the resistor is across the coil, and the resistor's share is proportional to the current. After three time constants the coil has almost stopped mattering, which is why steady state analysis ignores it.
How long until the current reaches three quarters of its final value
A $12.0\ \mathrm{V}$ source of negligible internal resistance is switched at $t = 0$ onto a $30.0\ \mathrm{\Omega}$ resistor in series with a $0.180\ \mathrm{H}$ coil of negligible resistance. Find the time constant, the final current, the current at $t = 3.00\ \mathrm{ms}$, and the time at which the current reaches $0.300\ \mathrm{A}$.
the coil and the source have no resistance of their own
Find
the time constant, the final current, $I$ at $3.00\ \mathrm{ms}$, and the time to reach $0.300\ \mathrm{A}$
Solution
The time constant and the final current are pinned down first and the exponent is then kept in time constants rather than in seconds, because both the forwards and the backwards question reduce to a pure number that way, and milliseconds inside an exponent are the usual way this goes wrong.
Independent check by landmarks rather than by recomputation: seventy five per cent must fall between one time constant, which gives sixty three per cent, and two, which gives eighty six. The answer $8.32\ \mathrm{ms}$ is $1.39$ time constants, comfortably inside that window.
The exponential was used forwards once and backwards once, and the backwards use needed one logarithm.
The forwards and backwards uses of the same formula are the two things this circuit is ever asked for, and the only new skill is isolating the exponential before taking a logarithm.
The spark at the switch: why twelve volts makes forty thousand
The circuit above has reached its steady current of $0.400\ \mathrm{A}$ when the switch is opened. As the contacts separate, the gap between them behaves for a moment like a resistance of $1.00\times10^{5}\ \mathrm{\Omega}$ in series with the coil. Find the new time constant and the voltage that appears across the gap at the first instant.
Given
$L = 0.180\ \mathrm{H}$ carrying $I_0 = 0.400\ \mathrm{A}$ at the moment of opening
the gap behaves as $R_{\rm gap} = 1.00\times10^{5}\ \mathrm{\Omega}$
the source is disconnected by the same action
$I(t) = I_0e^{-t/\tau}$ for a decaying loop
Find
the new time constant, and the voltage across the gap at the first instant
Solution
Ohm's law on the gap is chosen over differentiating the exponential, because the current at the first instant is known exactly while the derivative route needs the new time constant first; the derivative is then run anyway, as an independent check rather than as the main road.
Decide what has changed and what has not
$$I(0^{+}) = I_0 = 0.400\ \mathrm{A}$$
the current through an inductor cannot jump, so whatever the switch does, the current just after opening is the current just before it; this single fact is what makes the answer large
Energy check, independent of both routes above: the coil held $\tfrac12LI_0^{2} = 14.4\ \mathrm{mJ}$, and the gap dissipates $I_0^{2}R\tau'/2 = 14.4\ \mathrm{mJ}$ over the decay. The energy in the spark is exactly the energy that was in the field, which is where it had to come from.
Two independent routes to the same voltage, precisely because a number four thousand times the source voltage deserves a second opinion.
This is the hook, answered. Nothing supplied those volts except the coil's refusal to let its current change, and the reason it is a spark rather than a bang is the last line of the verification: fourteen millijoules is a very small amount of energy delivered very quickly. It is also why real circuits with coils in them carry a diode across the coil, giving the current a harmless path to die in.
Checkpoint
§13.4 — time constant and current at one time constant●●○○○
Thirty seconds, two short steps. A $20.0\ \mathrm{V}$ source is switched onto a $50.0\ \mathrm{\Omega}$ resistor in series with a $0.250\ \mathrm{H}$ coil, with the current starting from zero.
(b) What is the current at $t = 5.00\ \mathrm{ms}$?
Hint 1/4
Compare the time asked about with the time constant before doing anything else; if they are equal, the answer is a number you already know.
Hint 2/4
$\tau = L/R$ and $I = (V_0/R)(1-e^{-t/\tau})$, and at $t = \tau$ the bracket is $1-e^{-1} = 0.632$.
Hint 3/4
With $L = 0.250\ \mathrm{H}$ and $R = 50.0\ \mathrm{\Omega}$, $\tau = 5.00\ \mathrm{ms}$, which is exactly the time asked about, and $V_0/R = 0.400\ \mathrm{A}$.
Hint 4/4
$I = (0.400)(0.632) = 0.253\ \mathrm{A}$.
Show solution
The stated time is exactly one time constant, so we read the landmark $1-e^{-1}=0.632$ off the curve instead of reaching for a calculator.
Independent check on the units of $\tau$: henrys over ohms is volt seconds per amp divided by volts per amp, which leaves seconds, so the five is milliseconds and not something else.
⚠ Writing the time constant as $RL$
the capacitor case is $RC$, a product, and the pattern is copied across without checking
wrong$$\tau = RL$$
right$$\tau = \dfrac{L}{R}$$
⚠ Letting the current jump at the switch
a switch feels instantaneous, and everything else in the circuit does change instantly
right$$e^{-t/\tau} = 1-\frac{IR}{V_0}\ \Rightarrow\ t = -\tau\ln\!\left(1-\frac{IR}{V_0}\right)$$
13.5The LC circuit: charge that sloshes
A charged capacitor emptying into a coil overshoots, recharges backwards, and keeps going at a frequency the two components fix between them.
A coil stores energy and so does a capacitor, and neither of them can throw it away. Wire the two together and the energy has nowhere to go but back and forth.
TheoremResult 13.5: free oscillation of a capacitor and a coil
Conditions
one loop containing only $L$ and $C$, with no source and, for the first two lines, no resistance
the oscillation starts from a stated initial condition, usually a charged capacitor at rest
the third line applies when a resistance $R$ is present and is small enough that $R^{2} < 4L/C$; at $R^{2} = 4L/C$ the circuit stops oscillating altogether
no energy leaves the loop except through that resistance
The charge on the capacitor swings like a pendulum: one over the square root of the inductance times the capacitance gives the angular frequency in radians per second. Make either component larger and the swing gets slower, because a bigger capacitor takes longer to empty and a bigger coil resists the emptying more. Add resistance and the swing slows a little and dies away; add enough and it never swings at all.
Proof
Take the loop rule round the single loop, with $q$ the charge on the capacitor and $I = dq/dt$ the current flowing into it: $\dfrac{q}{C} + L\dfrac{dI}{dt} = 0$.
Replace the current by the derivative of the charge: $L\dfrac{d^{2}q}{dt^{2}} + \dfrac{q}{C} = 0$, or $\dfrac{d^{2}q}{dt^{2}} = -\dfrac{1}{LC}q$.
That is the equation of simple harmonic motion, letter for letter. A mass on a spring obeys $d^{2}x/dt^{2} = -(k/m)x$ with $\omega = \sqrt{k/m}$, so here $\omega_0^{2} = 1/(LC)$ and the solution is a cosine.
The translation is worth writing down once: the charge plays the part of the displacement, the current the part of the velocity, the inductance the part of the mass, and one over the capacitance the part of the spring constant. Every result you know about a mass on a spring is now a result about this circuit.
The energy follows from that mapping. The capacitor holds $q^{2}/2C$, playing the part of $\tfrac12kx^{2}$, and the coil holds $\tfrac12LI^{2}$, playing the part of $\tfrac12mv^{2}$. Their sum is constant, and each is largest exactly when the other is zero.
With a resistance in the loop the equation gains a $R\,dq/dt$ term and becomes the damped oscillator, whose frequency is lowered to $\omega'$ as stated. The change is small whenever $R$ is small: in the worked example below, fifteen ohms shifts the frequency by less than one per cent.
One full cycle of a lossless oscillation. The $\textcolor{#1f6feb}{\text{capacitor's energy}}$ and the $\textcolor{#d1690a}{\text{coil's energy}}$ each rise and fall twice per cycle, in opposite phase, and their sum never moves.
Looks like this, but is not
A bigger capacitor holds more energy, so it should drive the oscillation harder and make it faster. More charge means more push, and more push usually means quicker.
The capacitance sits under a square root in the denominator, so a bigger capacitor makes the oscillation slower, not faster. More charge does mean more energy, but it also means more charge to move, and the second effect wins. The same trap works for the coil, where the extra inductance both stores more and resists more. The mass on a spring makes it obvious: loading more mass on does not speed the oscillation up.
charge on the capacitor
energy in the capacitor
energy in the coil
current
$6.00\ \mathrm{\mu C}$
$2.25\ \mathrm{\mu J}$
$0$
$0$
$5.20\ \mathrm{\mu C}$
$1.69\ \mathrm{\mu J}$
$0.563\ \mathrm{\mu J}$
$6.71\ \mathrm{mA}$
$4.24\ \mathrm{\mu C}$
$1.13\ \mathrm{\mu J}$
$1.13\ \mathrm{\mu J}$
$9.49\ \mathrm{mA}$
$3.00\ \mathrm{\mu C}$
$0.563\ \mathrm{\mu J}$
$1.69\ \mathrm{\mu J}$
$11.6\ \mathrm{mA}$
$0$
$0$
$2.25\ \mathrm{\mu J}$
$13.4\ \mathrm{mA}$
The two energy columns add to $2.25\ \mathrm{\mu J}$ in every row, which is the whole content of the oscillation. The even split happens when the charge is down to seventy per cent of its peak, not to half of it, because the energy depends on the square of the charge.
A charged capacitor let loose into a coil
A $8.00\ \mathrm{\mu F}$ capacitor is charged to $6.00\ \mathrm{\mu C}$ and then connected at $t = 0$ across a $25.0\ \mathrm{mH}$ coil of negligible resistance. Find the angular frequency, the frequency in hertz, the period, the peak current, and the total energy, checking the energy two ways.
$Q_0 = 6.00\times10^{-6}\ \mathrm{C}$ at $t = 0$, with the current zero at that instant
no resistance in the loop
Find
$\omega_0$, $f$, $T$, $I_0$, and the total energy by two routes
Solution
The angular frequency is built from the components alone before anything else, because every other quantity asked for hangs off it; the energy is then taken at the two ends of the swing, where one term is the whole of it, rather than at a general instant where both terms would have to be carried.
a quarter of a cycle later the capacitor is empty and all of it is in the coil, so this term must come to the same number
Answer $$\boxed{\ \omega_0 = 2.24\times10^{3}\ \mathrm{rad/s},\ f = 356\ \mathrm{Hz},\ T = 2.81\ \mathrm{ms},\ I_0 = 13.4\ \mathrm{mA},\ U = 2.25\ \mathrm{\mu J}\ }$$
Check
The two energy routes agree, and they are genuinely independent: the first used $Q_0$ and $C$, the second used $L$ and a current that came from $\omega_0$. An error anywhere in the frequency would show up as a mismatch here.
One square root, one multiplication and two energy substitutions. The energy check was free because both quantities were already on the page.
Every LC question is some subset of these five numbers, and the energy check at the end costs one line and catches almost every arithmetic slip that matters.
What fifteen ohms does to the same circuit
The same coil and capacitor, $L = 25.0\ \mathrm{mH}$ and $C = 8.00\ \mathrm{\mu F}$, are now used with the coil's winding resistance of $15.0\ \mathrm{\Omega}$ taken into account. Find the oscillation frequency and the percentage by which it differs from the undamped value, then find the resistance at which the circuit would stop oscillating altogether.
the undamped value from the previous example is $\omega_0 = 2.24\times10^{3}\ \mathrm{rad/s}$
Find
$\omega'$, its percentage shift from $\omega_0$, and the resistance at which oscillation ceases
Solution
The two terms under the root are evaluated separately before they are subtracted, because they differ by a factor of fifty and a single combined line would hide whether the resistance matters at all.
Compare the two terms under the root before combining them
putting the two terms side by side before subtracting shows at once that the correction is under two per cent, so a large shift in the answer would be a signal of arithmetic trouble
using the unrounded values for a difference of close numbers, because rounding both to three figures first would destroy the very quantity being asked for
the oscillation ends when the term under the root reaches zero, since beyond that the frequency would be imaginary and the charge decays without ever crossing zero
Independent check on $R_c$ by units: $\sqrt{L/C}$ is $\sqrt{\mathrm{H/F}}$, and since $\mathrm{H} = \Omega\,\mathrm{s}$ and $\mathrm{F} = \mathrm{s}/\Omega$, the ratio is $\Omega^{2}$ and its root is an ohm, as a critical resistance must be.
Two squarings, one subtraction and two roots. The comparison of the two terms before subtracting was the step that made the answer easy to trust.
The general lesson: in a lightly damped oscillator the frequency barely moves, so if a problem asks only for a frequency and the resistance is small, the undamped formula is usually accurate enough. What the resistance really changes is the amplitude, which shrinks a little every cycle.
Checkpoint
§13.5 — natural frequency of an LC pair●●○○○
Thirty seconds, one square root. A $10.0\ \mathrm{mH}$ coil is connected across a $100\ \mathrm{\mu F}$ capacitor, with no resistance worth counting.
Given
$L = 10.0\times10^{-3}\ \mathrm{H}$
$C = 100\times10^{-6}\ \mathrm{F}$
$\omega_0 = 1/\sqrt{LC}$
Find
(a) What is the angular frequency of the oscillation?
(b) What is the frequency in hertz?
Hint 1/4
Two parts, two different quantities. Get the radians per second first, because that is what the formula produces, and convert afterwards.
Hint 2/4
$\omega_0 = 1/\sqrt{LC}$ and $f = \omega_0/2\pi$.
Hint 3/4
With $L = 10.0\ \mathrm{mH}$ and $C = 100\ \mathrm{\mu F}$ the product is $LC = 1.00\times10^{-6}\ \mathrm{s^{2}}$, whose square root is $1.00\times10^{-3}\ \mathrm{s}$.
Hint 4/4
$\omega_0 = 1.00\times10^{3}\ \mathrm{rad/s}$ and $f = 159\ \mathrm{Hz}$.
Show solution
The product $LC$ is formed first and left as a clean power of ten, so the square root can be taken by eye rather than trusted to a keystroke.
Independent check by period: $T = 1/f = 6.28\ \mathrm{ms}$, and $2\pi/\omega_0$ is also $6.28\ \mathrm{ms}$, which ties the two answers together through a quantity neither of them used directly.
⚠ Forgetting the square root in the natural frequency
the formula is short and the root is easy to lose when it is written on one line in a hurry
wrong$$\omega_0 = \dfrac{1}{LC}$$
right$$\omega_0 = \dfrac{1}{\sqrt{LC}}$$
⚠ Reporting the angular frequency as though it were in hertz
both are called frequency in speech, and the formula gives one of them while the question usually wants the other
wrong$$f = \dfrac{1}{\sqrt{LC}}$$
right$$f = \dfrac{1}{2\pi\sqrt{LC}}$$
⚠ Splitting the energy evenly at half the peak charge
half the charge sounds like half the energy, because the relationship between them is squared and easy to forget
A coil and a capacitor each limit an alternating current by an amount that depends on frequency, and neither of them consumes any average power.
The last circuit oscillated on its own. Now drive one with a source that oscillates on its own timetable, and take the components one at a time before putting them together.
RuleRule 13.6: the three elements alone across an alternating source
Conditions
a sinusoidal source of angular frequency $\omega = 2\pi f$, with one element alone across it
peak and root mean square values are related by $V_{\rm rms} = V_0/\sqrt2$, and the relations below hold for either pair as long as the same kind is used on both sides
the components are ideal: the coil has no resistance and the capacitor no leakage
the circuit has settled, so the current is a steady sine wave and no switch on behaviour is left
Each element relates its voltage to its current by a number of ohms, and for the coil and the capacitor that number depends on the frequency in opposite ways: the coil blocks fast changes and passes slow ones, the capacitor does the reverse. The other half of each line is the timing. In a resistor the voltage peaks with the current, in a coil it peaks a quarter of a cycle earlier, and in a capacitor a quarter of a cycle later.
Proof
Take the coil first, and let the current through it be $I = I_0\sin\omega t$, since in a series circuit the current is the quantity common to everything.
The voltage across it is $v_L = L\,dI/dt = \omega LI_0\cos\omega t$. Two things fall out at once: the peak voltage is $\omega LI_0$, so the ohms are $\omega L$, and the cosine peaks a quarter of a cycle before the sine, so the voltage leads.
Now the capacitor, driven the same way. Its charge is $q = Cv_C$, so $I = dq/dt = C\,dv_C/dt$, which rearranges to $v_C = \dfrac{1}{C}\displaystyle\int I\,dt = -\dfrac{I_0}{\omega C}\cos\omega t$.
The peak voltage is $I_0/(\omega C)$, so the ohms are $1/(\omega C)$, and the minus sign in front of the cosine puts the voltage a quarter of a cycle behind the current instead of ahead of it.
The average power follows from the timing alone. For the coil, $p = vI = \omega LI_0^{2}\sin\omega t\cos\omega t = \tfrac12\omega LI_0^{2}\sin2\omega t$, whose average over a cycle is zero: energy goes into the field for a quarter of a cycle and comes back out in the next.
The same argument, with the same trigonometric identity, gives zero for the capacitor. Only the resistor, where $v$ and $I$ peak together and their product is never negative, has a nonzero average, and it is $I_{\rm rms}^{2}R$.
The same coil, $0.150\ \mathrm{H}$, and the same capacitor, $4.00\ \mathrm{\mu F}$, across the whole audio range. The $\textcolor{#1f6feb}{\text{coil's reactance}}$ climbs in a straight line and the $\textcolor{#d1690a}{\text{capacitor's}}$ falls away as one over the frequency; they are equal at only one point.
Looks like this, but is not
Reactance is measured in ohms, so a coil with $47\ \mathrm{\Omega}$ of reactance heats up like a $47\ \mathrm{\Omega}$ resistor. It limits the current by the same amount, so it should cost the same in power.
The average power in an ideal coil is exactly zero, however many ohms of reactance it has. The reason is timing rather than size: the voltage and the current are a quarter of a cycle apart, so their product is positive for one quarter and negative for the next, and the energy that goes into the field comes straight back out. A resistor's voltage and current peak together, so its product is never negative and there is nothing to give back.
frequency
reactance of the coil
reactance of the capacitor
which one passes more current
$50.0\ \mathrm{Hz}$
$47.1\ \mathrm{\Omega}$
$796\ \mathrm{\Omega}$
the coil, by about seventeen times
$205\ \mathrm{Hz}$
$193\ \mathrm{\Omega}$
$194\ \mathrm{\Omega}$
neither; they are level here
$1.00\ \mathrm{kHz}$
$942\ \mathrm{\Omega}$
$39.8\ \mathrm{\Omega}$
the capacitor, by about twenty four times
$5.00\ \mathrm{kHz}$
$4.71\times10^{3}\ \mathrm{\Omega}$
$7.96\ \mathrm{\Omega}$
the capacitor, by nearly six hundred times
A hundredfold rise in frequency multiplies the coil's ohms by a hundred and divides the capacitor's by a hundred, so their ratio moves by ten thousand. This is why the same pair of components can be a treble filter or a bass filter depending only on which one the signal is asked to pass through.
A coil and a capacitor, each alone on the mains
A $0.150\ \mathrm{H}$ coil of negligible resistance is connected across a $220\ \mathrm{V}$ root mean square supply at $50.0\ \mathrm{Hz}$. Find its reactance, the root mean square current and the average power. Then replace it with a $4.00\ \mathrm{\mu F}$ capacitor and answer the same three questions.
Given
$V_{\rm rms} = 220\ \mathrm{V}$ at $f = 50.0\ \mathrm{Hz}$
first case: $L = 0.150\ \mathrm{H}$, no resistance
second case: $C = 4.00\times10^{-6}\ \mathrm{F}$, no leakage
Find
the reactance, current and average power in each case
Solution
The angular frequency is converted once at the top and reused for both components, because the whole of this question is two reactance formulas and the only real risk is losing the factor of $2\pi$ halfway down.
Convert the frequency once, before anything else
$$\omega = 2\pi f = 2\pi(50.0) = 314\ \mathrm{rad/s}$$
every reactance formula wants radians per second, and doing this conversion on its own line is the cheapest possible insurance against losing the factor of $2\pi$
The coil
$$X_L = \omega L = (314)(0.150) = 47.1\ \mathrm{\Omega}$$
at mains frequency a coil of this size is a mild obstruction, which is why for mains use are much larger than this one
root mean square voltage divided by ohms gives root mean square current, and mixing a peak with a root mean square here is the standard way to be out by a factor of $\sqrt2$
$$P_{\rm avg} = 0$$
the voltage and the current are a quarter of a cycle apart, so the energy that flows in during one quarter flows back out during the next
seventeen times less current than the coil allowed, from components of ordinary size, purely because of the frequency
$$P_{\rm avg} = 0$$
same argument, opposite sign of the phase shift; a quarter cycle either way still averages to nothing
Answer $$\boxed{\ X_L = 47.1\ \mathrm{\Omega},\ I = 4.67\ \mathrm{A},\ P = 0;\qquad X_C = 796\ \mathrm{\Omega},\ I = 0.276\ \mathrm{A},\ P = 0\ }$$
Check
Independent check by the product of the two reactances: $X_LX_C = (\omega L)/(\omega C) = L/C = 0.150/4.00\times10^{-6} = 3.75\times10^{4}$, a number that does not contain the frequency at all. The two computed values multiply to $(47.1)(796) = 3.75\times10^{4}$, which tests both of them at once.
One frequency conversion, two substitutions and two divisions. The two power answers cost nothing.
Zero average power does not mean zero current, and the coil here is drawing nearly five amps while consuming nothing. That current is real, it heats the wiring, and it is why the electricity supply industry cares about power factor, which turns up at the end of the next concept.
The same two components at five kilohertz, and the frequency where they meet
Take the same $0.150\ \mathrm{H}$ coil and $4.00\ \mathrm{\mu F}$ capacitor to $5.00\ \mathrm{kHz}$ and find both reactances. Then find the one frequency at which the two are equal, and the value they share there.
both reactances at $5.00\ \mathrm{kHz}$, and the frequency at which they are equal
Solution
The crossing is found by setting the two reactances equal and solving symbolically rather than by trying frequencies, because the algebra lands on $1/\sqrt{LC}$, a formula already in hand from the previous concept.
setting the two expressions equal and solving symbolically, rather than searching numerically, because the answer turns out to be a formula already met in the previous concept
substituting back with the unrounded angular frequency rather than the rounded frequency in hertz, because rounding to three figures first and then multiplying would have shifted the last digit
Independent check on the crossing value using the frequency free product from the previous example: $X_LX_C = L/C = 3.75\times10^{4}$ always, so when the two are equal each must be $\sqrt{3.75\times10^{4}} = 194\ \mathrm{\Omega}$, reached without using the frequency at all.
Two proportional scalings that needed no calculator, then one symbolic solve.
That crossing frequency is the natural frequency of the previous concept, arrived at from a completely different direction. It is not a coincidence, and it is the reason the next concept has a resonance in it.
Checkpoint
§13.6 — reactance and current for a capacitor alone●●○○○
Thirty seconds, two steps. A $2.00\ \mathrm{\mu F}$ capacitor is connected alone across a $120\ \mathrm{V}$ root mean square supply at $60.0\ \mathrm{Hz}$.
Given
$C = 2.00\times10^{-6}\ \mathrm{F}$
$V_{\rm rms} = 120\ \mathrm{V}$ at $f = 60.0\ \mathrm{Hz}$
$X_C = 1/(\omega C)$ with $\omega = 2\pi f$
Find
(a) What is the capacitor's reactance?
(b) What root mean square current flows, and what average power does the capacitor consume?
Hint 1/4
Two of the three answers need arithmetic and one does not. Decide which is which before starting.
Hint 2/4
$X_C = 1/(2\pi fC)$, then $I_{\rm rms} = V_{\rm rms}/X_C$, and the average power in an ideal capacitor is zero whatever the numbers.
Hint 3/4
With $f = 60.0\ \mathrm{Hz}$, $\omega = 377\ \mathrm{rad/s}$, and with $C = 2.00\times10^{-6}\ \mathrm{F}$ the product $\omega C = 7.54\times10^{-4}$, while $V_{\rm rms} = 120\ \mathrm{V}$.
The average power is settled from the phase relation and not by computing $I^{2}X_C$: a reactance stores and returns, so no arithmetic on it can produce a dissipated watt.
Angular frequency, then reactance
$$\omega = 2\pi(60.0) = 377\ \mathrm{rad/s}$$
this number appears in almost every mains problem and is worth recognising on sight
Independent check by charge: the peak charge is $CV_0 = (2.00\times10^{-6})(120\sqrt2) = 3.39\times10^{-4}\ \mathrm{C}$, and the peak current $\omega Q_0 = (377)(3.39\times10^{-4}) = 0.128\ \mathrm{A}$, whose root mean square value is $0.128/\sqrt2 = 0.0905\ \mathrm{A}$.
⚠ Using the frequency in hertz where the angular frequency belongs
the supply is quoted in hertz and the formula is written in omega, and the two are only ever six steps apart in a hurried line of algebra
wrong$$X_L = fL$$
right$$X_L = \omega L = 2\pi fL$$
⚠ Writing the the right way up
the inductive one is a product, and symmetry suggests the capacitive one should be too
wrong$$X_C = \omega C$$
right$$X_C = \dfrac{1}{\omega C}$$
⚠ Charging the coil or the capacitor for power
the units are ohms and every other quantity measured in ohms so far has dissipated heat
wrong$$P_{\rm avg} = I_{\rm rms}^{2}X_L$$
right$$P_{\rm avg} = 0\ \text{for an ideal}\ L\ \text{or}\ C;\quad P_{\rm avg} = I_{\rm rms}^{2}R\ \text{only in a resistance}$$
13.7The series LRC circuit and resonance
Three elements in one loop combine at right angles rather than by addition, and at one frequency the two reactive parts cancel.
Each element alone is settled. Put all three in one loop and the only difficulty left is that their voltages peak at different instants, so they cannot simply be added.
TheoremResult 13.7: impedance, phase and power of a series LRC circuit
Conditions
$R$, $L$ and $C$ in a single series loop across a sinusoidal source of angular frequency $\omega$
the circuit has settled into steady oscillation at the source's frequency, not at its own
the current is the same everywhere in the loop, which is what makes it the natural reference for every phase
the power expression assumes $R$ is the only resistance present, including any winding resistance of the coil
Subtract the two reactances, square what is left, add the square of the resistance and take the root: that is the ohms the source sees. The angle whose tangent is the reactance difference over the resistance tells you how far the source voltage runs ahead of the current. The power is the current squared times the resistance and nothing else, because the resistance is the only place in the loop where energy stops coming back.
Proof
The current is common to all three elements, so make it the reference and ask where each voltage sits relative to it.
The resistor's voltage peaks with the current. The coil's peaks a quarter of a cycle earlier and the capacitor's a quarter of a cycle later, so those two are half a cycle apart from each other, which means they are in direct opposition and subtract.
Represent each voltage by an arrow whose length is its peak value and whose direction shows its timing. The resistor's arrow lies along the current, the coil's points a quarter turn one way and the capacitor's a quarter turn the other.
The source voltage is the sum of the three, added as arrows: the two reactive ones cancel along their shared line, leaving $|V_L-V_C|$ at right angles to $V_R$.
By Pythagoras, $V = \sqrt{V_R^{2}+(V_L-V_C)^{2}} = I\sqrt{R^{2}+(X_L-X_C)^{2}}$, which defines the impedance, and the angle of the resultant above the current is $\phi$ with $\tan\phi = (X_L-X_C)/R$.
For the power, only the component of the source voltage along the current does any net work, and that component is $V\cos\phi$. Since $V\cos\phi = IR$ from the triangle, the two forms of the power expression are the same statement, and both say the energy goes into the resistor.
Resonance is the case $X_L = X_C$, which happens at $\omega_0 = 1/\sqrt{LC}$. There the reactance difference vanishes, $Z$ falls to its smallest possible value $R$, the current is at its largest, and $\phi = 0$.
Why the three ohms do not add. The $\textcolor{#8250df}{\text{resistance}}$ and the $\textcolor{#d1690a}{\text{net reactance}}$ meet at a right angle, so the $\textcolor{#1f6feb}{\text{impedance}}$ is the hypotenuse and the phase angle is built into the same triangle. The numbers are those of the worked example below.
Looks like this, but is not
The voltages across the three elements must add up to the source voltage. A voltmeter reads $66.0\ \mathrm{V}$ across the resistor, $39.8\ \mathrm{V}$ across the coil and $140\ \mathrm{V}$ across the capacitor, so the source must be supplying $246\ \mathrm{V}$.
The source in that circuit is $120\ \mathrm{V}$. Each meter reads a genuine root mean square voltage, but the three peak at different instants, so at no moment are all three at the values shown. Combining them properly means subtracting the two reactive readings first and then using Pythagoras, which gives exactly the source value back. Notice also that one reading is larger than the source on its own, which is normal here and impossible in a direct current circuit.
frequency
impedance
current
phase angle
average power
$60.0\ \mathrm{Hz}$
$455\ \mathrm{\Omega}$
$0.264\ \mathrm{A}$
$-56.6^{\circ}$
$17.4\ \mathrm{W}$
$90.0\ \mathrm{Hz}$
$281\ \mathrm{\Omega}$
$0.428\ \mathrm{A}$
$-27.0^{\circ}$
$45.7\ \mathrm{W}$
$113\ \mathrm{Hz}$
$250\ \mathrm{\Omega}$
$0.480\ \mathrm{A}$
$0^{\circ}$
$57.6\ \mathrm{W}$
$150\ \mathrm{Hz}$
$299\ \mathrm{\Omega}$
$0.401\ \mathrm{A}$
$+33.4^{\circ}$
$40.2\ \mathrm{W}$
$250\ \mathrm{Hz}$
$560\ \mathrm{\Omega}$
$0.214\ \mathrm{A}$
$+63.5^{\circ}$
$11.5\ \mathrm{W}$
The impedance has a floor at $250\ \mathrm{\Omega}$, which is the resistance, and it is reached at one frequency only. The phase angle passes through zero at that same frequency, changing sign from capacitive below it to inductive above it, and the power peaks there too, at more than three times its value at mains frequency.
A full series circuit at 60 hertz, including the voltage across each element
A $250\ \mathrm{\Omega}$ resistor, a $0.400\ \mathrm{H}$ coil and a $5.00\ \mathrm{\mu F}$ capacitor are connected in series across a $120\ \mathrm{V}$ root mean square supply at $60.0\ \mathrm{Hz}$. Find the impedance, the current, the phase angle, the average power, and the root mean square voltage across each element.
Given
$R = 250\ \mathrm{\Omega}$, $L = 0.400\ \mathrm{H}$, $C = 5.00\times10^{-6}\ \mathrm{F}$, in series
$V_{\rm rms} = 120\ \mathrm{V}$ at $f = 60.0\ \mathrm{Hz}$
the coil has no resistance of its own
Find
$Z$, $I_{\rm rms}$, $\phi$, $P_{\rm avg}$, and $V_R$, $V_L$, $V_C$
Solution
The reactance difference is carried with its sign all the way down instead of being squared early, because the phase angle needs that sign even though the impedance is about to throw it away.
The two reactances, before anything is combined
$$\omega = 2\pi(60.0) = 377\ \mathrm{rad/s}$$
one conversion, done once, used four times below
$$X_L = \omega L = (377)(0.400) = 151\ \mathrm{\Omega}$$
the sum of the three ohm values would have been $932\ \mathrm{\Omega}$, so this is not a small correction to a naive addition; it is a different calculation
and note that the capacitor's reading exceeds the supply, which is allowed here and would be impossible with a battery
Answer $$\boxed{\ Z = 455\ \mathrm{\Omega},\ I = 0.264\ \mathrm{A},\ \phi = -56.6^{\circ},\ P = 17.4\ \mathrm{W};\ \ V_R = 66.0,\ V_L = 39.8,\ V_C = 140\ \mathrm{V}\ }$$
Check
Independent check that recombines the element voltages instead of repeating any earlier line: $\sqrt{V_R^{2}+(V_L-V_C)^{2}} = \sqrt{(66.0)^{2}+(-100)^{2}} = 120\ \mathrm{V}$, which is the supply. A second check on the power from the other formula: $V_{\rm rms}I_{\rm rms}\cos\phi = (120)(0.264)(0.550) = 17.4\ \mathrm{W}$.
Two reactances, one Pythagoras, one arctangent and three multiplications. The two verifications cost two more lines and are worth every one.
The number to carry away is $455\ \mathrm{\Omega}$ against a naive $932\ \mathrm{\Omega}$. Adding the ohms would have halved the current, and the whole of that difference comes from the two reactances working against each other rather than together.
The same circuit at resonance, where the reactances cancel
For the same $250\ \mathrm{\Omega}$, $0.400\ \mathrm{H}$ and $5.00\ \mathrm{\mu F}$ series circuit on the same $120\ \mathrm{V}$ supply, find the frequency at which the current is largest, and at that frequency find the impedance, the current, the average power, and the voltage across the coil.
$V_{\rm rms} = 120\ \mathrm{V}$, frequency adjustable
resonance is the condition $X_L = X_C$
Find
the , and $Z$, $I$, $P$ and $V_L$ there
Solution
Resonance is located from $X_L = X_C$ rather than by maximising the current expression: with $R$ fixed, the smallest impedance is exactly the vanishing of the reactance difference, so no calculus is needed.
the reactances are not zero at resonance; they are equal, which is a completely different statement
$$V_L = IX_L = (0.480)(283) = 136\ \mathrm{V}$$
larger than the $120\ \mathrm{V}$ supply, and the capacitor carries the same $136\ \mathrm{V}$ in exact opposition, so the two cancel as far as the loop is concerned
Answer $$\boxed{\ f_0 = 113\ \mathrm{Hz},\ Z = 250\ \mathrm{\Omega},\ I = 0.480\ \mathrm{A},\ P = 57.6\ \mathrm{W},\ V_L = 136\ \mathrm{V}\ }$$
Check
Independent check on $V_L$ through the capacitor instead: $X_C = 1/(\omega_0C) = 1/((707)(5.00\times10^{-6})) = 283\ \mathrm{\Omega}$, so $V_C = 136\ \mathrm{V}$ as well, and $V_L-V_C = 0$ leaves $V_R = (0.480)(250) = 120\ \mathrm{V}$, exactly the supply.
One square root fixes the frequency and every other answer follows in a single line, which is what makes resonance questions quick once the condition is recognised.
Two things to carry away. The current is at its largest here and the power more than triples, and yet the coil and the capacitor are each carrying more volts than the supply. Resonance does not remove the reactances; it makes them cancel each other while both stay large.
Checkpoint
§13.7 — impedance and power from three given ohm values●●○○○
Thirty seconds, deliberately clean numbers. In a series circuit driven at some fixed frequency, the resistance is $30.0\ \mathrm{\Omega}$, the is $80.0\ \mathrm{\Omega}$ and the capacitive reactance is $40.0\ \mathrm{\Omega}$, across a $100\ \mathrm{V}$ root mean square supply.
Given
$R = 30.0\ \mathrm{\Omega}$
$X_L = 80.0\ \mathrm{\Omega}$ and $X_C = 40.0\ \mathrm{\Omega}$
$V_{\rm rms} = 100\ \mathrm{V}$
Find
(a) What is the impedance and the current?
(b) What is the average power delivered?
Hint 1/4
The reactances are given, so there is no frequency work at all. The only decision is how the three ohm values combine.
Hint 2/4
$Z = \sqrt{R^{2}+(X_L-X_C)^{2}}$, then $I = V/Z$, and $P_{\rm avg} = I^{2}R$ using the resistance alone.
Hint 3/4
With $R = 30.0\ \mathrm{\Omega}$, $X_L = 80.0\ \mathrm{\Omega}$ and $X_C = 40.0\ \mathrm{\Omega}$, the difference is $40.0\ \mathrm{\Omega}$ and the supply is $100\ \mathrm{V}$.
The reactances are subtracted before anything is squared. Adding all three ohm values is the tempting alternative and gives $150\ \mathrm{\Omega}$ where the right answer is $50\ \mathrm{\Omega}$.
Combine the ohms correctly
$$X_L-X_C = 80.0-40.0 = 40.0\ \mathrm{\Omega}$$
the reactances oppose each other, so this subtraction comes before any squaring
the resistance is what the source pays for, so it is the right thing to multiply by here; the two must not be swapped
Answer $$\boxed{\ Z = 50.0\ \mathrm{\Omega},\quad I = 2.00\ \mathrm{A},\quad P_{\rm avg} = 120\ \mathrm{W}\ }$$
Check
Independent check through the phase angle: $\tan\phi = 40.0/30.0$ gives $\phi = 53.1^{\circ}$ and $\cos\phi = 0.600$, so $P = VI\cos\phi = (100)(2.00)(0.600) = 120\ \mathrm{W}$, matching by a route that never used the resistance directly.
⚠ Adding the three ohm values
series resistances add, and the pattern is applied to anything measured in ohms sitting in a series loop
wrong$$Z = R+X_L+X_C$$
right$$Z = \sqrt{R^{2}+(X_L-X_C)^{2}}$$
⚠ Using the impedance in the power formula
the impedance is the number that gave the current, so it feels like the number that should give the power
wrong$$P_{\rm avg} = I_{\rm rms}^{2}Z$$
right$$P_{\rm avg} = I_{\rm rms}^{2}R$$
⚠ Believing the reactances are zero at resonance
the word cancel is heard as vanish, and the impedance really does become just the resistance
Reading a switching circuit at the two instants that need no calculus
Any question containing the words just after the switch is closed, or a long time later. Both ends can be read off without solving anything, and doing so first tells you whether the exponential is needed at all.
Ask what cannot jump
The current through an inductor cannot jump, and the voltage across a capacitor cannot jump. Everything else in the circuit is allowed to change instantly.
At the first instant
An inductor that was carrying nothing carries nothing still, so it behaves as a break. A capacitor that was empty has no voltage across it, so it behaves as a wire.
A long time later
Nothing is changing, so an inductor has no voltage across it and behaves as a wire, while a capacitor carries no current and behaves as a break. The roles have swapped.
Fill in the rest of the circuit at each end
With the coil or capacitor replaced by a wire or a break, what is left is a resistor network you can already solve, and it gives the starting and finishing values of every current in the circuit.
Only now reach for the exponential
Any quantity in the circuit runs from its starting value to its finishing value as $\text{start}+(\text{finish}-\text{start})(1-e^{-t/\tau})$, with $\tau = L/R$ or $RC$. If the question asks only about the two ends, the exponential is never needed.
Where it goes wrong
Replacing the inductor by a wire at $t = 0$, which is the long time behaviour applied at the wrong end.
Using the source voltage divided by the resistance as the starting current in a switch on problem, when the starting current is zero.
Forgetting that the resistance in $\tau = L/R$ is the total resistance of the loop the current actually flows round, which changes when a switch changes the circuit.
Deciding which of the three frequency-like quantities a question wants
Every time a problem mentions a coil and a capacitor and a number of hertz, because three different quantities in this section are all called something like frequency and only one of them answers any given question.
Is anything oscillating on its own?
If the circuit contains no source and has been left to itself, the quantity wanted is $\omega_0 = 1/\sqrt{LC}$, the natural angular frequency, and it depends on the components only.
Is a source setting the pace?
If a source is driving the circuit, everything in it oscillates at the source's frequency $\omega = 2\pi f$, whatever the components would have preferred. Reactances are computed with this one.
Is the question about a peak, a maximum or a cancellation?
Then it is asking where the driving frequency equals the natural one. Set $\omega = \omega_0$, which is the same as $X_L = X_C$, and everything else simplifies.
Is nothing oscillating at all?
A switch, a battery and an exponential mean the quantity wanted is $\tau = L/R$, which is a time and not a frequency, and no factor of $2\pi$ belongs anywhere near it.
Write the chosen quantity down with its unit
Radians per second for $\omega$ and $\omega_0$, hertz for $f$, seconds for $\tau$. Almost every mistake in this section is one of these three quantities used in a formula built for another.
Where it goes wrong
Putting the source frequency into $1/\sqrt{LC}$, which is a formula that takes components and returns a frequency, not the other way round.
Reporting $\omega_0$ in hertz because the question asked for a frequency.
Treating $\tau = L/R$ as though a factor of $2\pi$ were missing from it.
Solving any driven series LRC circuit in six lines
A resistor, a coil and a capacitor in one loop across an alternating source, which is the standard examination question of this section and always yields to the same six lines in the same order.
Angular frequency
$\omega = 2\pi f$, written on its own line with its unit. Everything below depends on it.
The two reactances
$X_L = \omega L$ and $X_C = 1/(\omega C)$, both in ohms. Look at which is larger and predict the sign of the phase angle before computing it.
Impedance
$Z = \sqrt{R^{2}+(X_L-X_C)^{2}}$. Subtract before squaring, and keep the sign of the difference for the next line.
Current
$I_{\rm rms} = V_{\rm rms}/Z$. This one current then belongs to every element in the loop.
Phase angle
$\tan\phi = (X_L-X_C)/R$. Check the sign against the prediction from line two; a disagreement means an arithmetic slip, not a surprise.
Power, and the element voltages if asked
$P_{\rm avg} = I_{\rm rms}^{2}R$, using the resistance and never the impedance, and $V_R = IR$, $V_L = IX_L$, $V_C = IX_C$. Finish by checking $\sqrt{V_R^{2}+(V_L-V_C)^{2}}$ against the supply.
Where it goes wrong
Skipping line one and using $f$ in place of $\omega$, which multiplies both reactances wrongly by about six.
Adding the three ohm values instead of combining two of them at right angles to the third.
Using $Z$ in the power formula, which overstates the power by the factor $Z/R$.
Never doing the final check, which is the one line that would have caught any of the above.
A capacitor switched onto a battery through a resistor
A $12.0\ \mathrm{V}$ source is switched at $t = 0$ onto a $30.0\ \mathrm{\Omega}$ resistor in series with an uncharged $200\ \mathrm{\mu F}$ capacitor. Find the time constant and the current just after closing and long afterwards.
Given
$V_0 = 12.0\ \mathrm{V}$
$R = 30.0\ \mathrm{\Omega}$
$C = 200\times10^{-6}\ \mathrm{F}$, uncharged at $t = 0$
Find
the time constant and the current at both ends
Solution
The two ends of the curve are read off the physical state of the capacitor rather than by substituting $t=0$ and $t=\infty$ into an exponential, because the states are the part worth remembering and the exponential is not needed at either end.
here the resistance multiplies, so a larger resistor makes this circuit slower
The two ends
$$I(0^{+}) = \frac{V_0}{R} = 0.400\ \mathrm{A}$$
an empty capacitor has no voltage across it, so at the first instant it is indistinguishable from a piece of wire and the resistor sees the whole source
$$I(\infty) = 0$$
once the capacitor has charged to the source voltage there is nothing left to drive a current, so it behaves as a break
Check by energy: the source delivers charge until the capacitor holds $CV_0 = 2.40\ \mathrm{mC}$, and with no path for a steady current afterwards the final current has to be zero.
A coil switched onto the same battery through the same resistor
The same $12.0\ \mathrm{V}$ source is switched at $t = 0$ onto the same $30.0\ \mathrm{\Omega}$ resistor, now in series with a $0.180\ \mathrm{H}$ coil carrying no current. Find the time constant and the current just after closing and long afterwards.
Given
$V_0 = 12.0\ \mathrm{V}$
$R = 30.0\ \mathrm{\Omega}$
$L = 0.180\ \mathrm{H}$, no current at $t = 0$
Find
the time constant and the current at both ends
Solution
Deliberately the same route as the capacitor case: the point of the pair is that only two things change, where $R$ sits in the time constant and which end of the curve is the zero.
here the resistance divides, so a larger resistor makes this circuit faster, and the numbers have been chosen so that the two circuits share a time constant
The two ends
$$I(0^{+}) = 0$$
the current through a coil cannot jump, and it was zero before the switch closed, so at the first instant the coil behaves as a break
$$I(\infty) = \frac{V_0}{R} = 0.400\ \mathrm{A}$$
once nothing is changing the coil produces no voltage and behaves as a piece of wire, leaving the resistor alone with the source
Check by energy: at the end the coil holds $\tfrac12LI^{2} = 14.4\ \mathrm{mJ}$ and is passing a steady current, which is the exact reverse of the capacitor case, where the store was full and the current had stopped.
Same source, same resistor, same time constant, and the two currents run in opposite directions in time: the capacitor circuit starts at $0.400\ \mathrm{A}$ and ends at zero, while the coil circuit starts at zero and ends at $0.400\ \mathrm{A}$.
How to tell them apart
Ask which quantity is forbidden to jump. For a capacitor it is the voltage, so an empty capacitor starts as a wire and finishes as a break. For an inductor it is the current, so an empty inductor starts as a break and finishes as a wire. Then remember that the time constant is a product in one case and a quotient in the other: $RC$ grows with $R$ and $L/R$ shrinks with it.
Scaffolding comes off
The common skeleton
Write the angular frequency on its own line, in radians per second, converting from hertz if that is what was given.
Compute each reactance separately and label it, without combining anything yet.
Compare the two reactances and say in words which kind of circuit this is going to be, and therefore what sign the phase angle will have.
Combine at right angles: subtract the reactances, then use Pythagoras against the resistance to get the impedance.
Divide the supply voltage by the impedance to get the one current that flows through everything in the loop.
Get the phase angle from the reactance difference over the resistance, and check its sign against the prediction made in words.
Take the average power from the current and the resistance alone, then check the element voltages recombine to the supply.
1 · fully worked
A full series solve at 120 hertz
A $100\ \mathrm{\Omega}$ resistor, a $0.200\ \mathrm{H}$ coil and a $10.0\ \mathrm{\mu F}$ capacitor are in series across a $60.0\ \mathrm{V}$ root mean square supply at $120\ \mathrm{Hz}$. Find the impedance, the current, the phase angle and the average power.
$V_{\rm rms} = 60.0\ \mathrm{V}$ at $f = 120\ \mathrm{Hz}$
Find
$Z$, $I_{\rm rms}$, $\phi$ and $P_{\rm avg}$
Solution
We go down the method box in its order, frequency then reactances then their difference then the impedance, because every later line reuses an earlier one; jumping straight at $Z$ means computing $\omega$ twice and keeping neither.
Frequency and reactances
$$\omega = 2\pi(120) = 754\ \mathrm{rad/s}$$
the conversion is written out because every line below is proportional to it or to its reciprocal
when one side of the triangle is much smaller than the other, the hypotenuse barely exceeds the longer side, which is a useful thing to notice before trusting the arithmetic
resistance, not impedance, and the two differ by only two per cent here, which is exactly why using the wrong one would be hard to notice
Answer $$\boxed{\ Z = 102\ \mathrm{\Omega},\ I = 0.590\ \mathrm{A},\ \phi = +10.3^{\circ},\ P = 34.8\ \mathrm{W}\ }$$
Check
Independent check through the element voltages: $V_R = 59.0\ \mathrm{V}$, $V_L = 89.0\ \mathrm{V}$, $V_C = 78.3\ \mathrm{V}$, and $\sqrt{(59.0)^{2}+(89.0-78.3)^{2}} = 60.0\ \mathrm{V}$, the supply.
The circuit is close to its resonance at $113\ \mathrm{Hz}$, which is why the two reactances nearly cancel and the impedance is barely above the resistance.
2 · you write the reasoning
Easier than the last one: no impedance, no phase, no power, three lines. A coil of $0.120\ \mathrm{H}$ and a capacitor of $30.0\ \mathrm{\mu F}$ are each driven at $60.0\ \mathrm{Hz}$. All three lines below are correct. Before opening the model reasons, say in your own words why each one is allowed, and pay particular attention to what the third line lets you predict about a circuit containing both.
reasoning
The first line exists because every reactance formula is written in radians per second while every supply is quoted in hertz. Doing the conversion once, at the top, means the factor of $2\pi$ appears in exactly one place where it can be checked, instead of being carried in the head into two separate formulas.
reasoning
The second line is a multiplication because the coil's opposition grows with how fast the current is being asked to change, and at a fixed peak current a higher frequency means a steeper change. Nothing about the capacitor or about any circuit enters; this number belongs to the coil and the frequency alone.
reasoning
The third line is a reciprocal for the mirror image reason: a capacitor passes current more easily the faster it is being charged and discharged. The comparison is what matters for later. Here $X_C$ is about twice $X_L$, so a series circuit made of these two would be capacitive at this frequency, its phase angle would come out negative, and it would need a higher frequency to reach resonance.
3 · find the buried error
Harder than the last one: the same two components are now in a series circuit with a resistor, and the question asks for a power, so there are more places to go wrong. A $40.0\ \mathrm{\Omega}$ resistor, a $0.120\ \mathrm{H}$ coil and a $30.0\ \mathrm{\mu F}$ capacitor are in series across a $24.0\ \mathrm{V}$ root mean square supply at $60.0\ \mathrm{Hz}$. A student writes the five lines below and reports an average power of $6.36\ \mathrm{W}$. Find every faulty line; you are not told how many there are.
the two buried errors (2)
⚠ step 1
The inductive reactance was computed with the frequency in hertz instead of the angular frequency, leaving out the factor of $2\pi$.
The supply is quoted in hertz and the second line of the same solution uses the angular frequency correctly, so the slip is a single lapse rather than a misunderstanding, and it survives because $7.20\ \mathrm{\Omega}$ looks like a perfectly ordinary reactance.
right
$X_L = \omega L = (377)(0.120) = 45.2\ \mathrm{\Omega}$, which is more than six times the value used and changes the reactance difference from $-81.2\ \mathrm{\Omega}$ to $-43.2\ \mathrm{\Omega}$.
⚠ step 5
The average power was taken as the current squared times the impedance rather than times the resistance.
The impedance is the number that produced the current one line earlier, so it is the number sitting in front of the eye, and the formula $P = I^{2}R$ is remembered as current squared times whatever the ohms were.
right
$P_{\rm avg} = I_{\rm rms}^{2}R$, because only the resistance keeps any energy; the coil and the capacitor return everything they take. With the corrected current this gives $P = (0.408)^{2}(40.0) = 6.65\ \mathrm{W}$.
4 · the bare problem
§13.7 — a series circuit with no scaffolding●●●●○
No guidance this time. A $60.0\ \mathrm{\Omega}$ resistor, a $0.150\ \mathrm{H}$ coil and a $20.0\ \mathrm{\mu F}$ capacitor are connected in series across a $100\ \mathrm{V}$ root mean square supply at $50.0\ \mathrm{Hz}$.
Given
$R = 60.0\ \mathrm{\Omega}$, $L = 0.150\ \mathrm{H}$, $C = 20.0\times10^{-6}\ \mathrm{F}$, in series
$V_{\rm rms} = 100\ \mathrm{V}$ at $f = 50.0\ \mathrm{Hz}$
Find
(a) Find the impedance and the root mean square current.
(b) Find the phase angle and say whether the current leads or lags.
(c) Find the average power delivered.
(d) Find the frequency at which this circuit would resonate.
Hint 1/4
Four parts, one circuit. The first three all follow from the two reactances, and the fourth needs neither the supply nor the resistance.
With $f = 50.0\ \mathrm{Hz}$, $\omega = 314\ \mathrm{rad/s}$; with $L = 0.150\ \mathrm{H}$, $X_L = 47.1\ \mathrm{\Omega}$; with $C = 20.0\ \mathrm{\mu F}$, $X_C = 159\ \mathrm{\Omega}$; and $R = 60.0\ \mathrm{\Omega}$ with a supply of $100\ \mathrm{V}$.
Hint 4/4
$Z = 127\ \mathrm{\Omega}$, $I = 0.787\ \mathrm{A}$, $\phi = -61.8^{\circ}$ with the current leading, $P = 37.2\ \mathrm{W}$, and $f_0 = 91.9\ \mathrm{Hz}$.
Show solution
The six lines run in the method box's order with no shortcuts, and the resonance is left to the end on purpose, because it needs only $L$ and $C$ and would be spoiled by dragging the supply values into it.
Frequency and the two reactances
$$\omega = 2\pi(50.0) = 314\ \mathrm{rad/s}$$
written first so that the factor of $2\pi$ enters the solution exactly once
the supply voltage and the resistance play no part in where the resonance sits, only in how sharp it is
Answer $$\boxed{\ Z = 127\ \mathrm{\Omega},\ I = 0.787\ \mathrm{A},\ \phi = -61.8^{\circ}\ \text{(current leads)},\ P = 37.2\ \mathrm{W},\ f_0 = 91.9\ \mathrm{Hz}\ }$$
Check
Independent check on the power through the other formula: $\cos(-61.8^{\circ}) = 0.472$, so $V_{\rm rms}I_{\rm rms}\cos\phi = (100)(0.787)(0.472) = 37.2\ \mathrm{W}$, agreeing without using the resistance directly.
Note that the resonant frequency came out above the supply frequency, which is exactly what the negative phase angle predicted: below resonance a series circuit is always capacitive.
Full exam-style question
Examination format: a coil switched on, its energy, and where the power goesexam format
A coil of self-inductance $0.500\ \mathrm{H}$ and negligible resistance is connected in series with a $25.0\ \mathrm{\Omega}$ resistor and a $50.0\ \mathrm{V}$ source of negligible internal resistance. The switch is closed at $t = 0$ with no current flowing. (a) Find the time constant and the final current. (b) Find the current at $t = 20.0\ \mathrm{ms}$. (c) Find the energy stored in the coil at that instant and the energy it will eventually hold. (d) At $t = 20.0\ \mathrm{ms}$, find the power supplied by the source, the power dissipated in the resistor, and the power going into the coil's field.
the time constant and final current, the current at $20.0\ \mathrm{ms}$, the stored energies, and the three powers at that instant
Solution
Part (d) is done by the energy balance rather than by differentiating, because the balance uses only numbers already computed in earlier parts, while differentiating introduces a fresh exponential and a fresh chance of error. The differentiated route is kept for the verification, where a disagreement would be informative rather than costly.
the energy depends on the square of the current, so at sixty three per cent of the final current the coil holds only forty per cent of its final energy
Independent check on the last figure from the coil's own formula rather than by subtraction: $dI/dt = (V_0/L)e^{-t/\tau} = (100)(0.368) = 36.8\ \mathrm{A/s}$, so $P_L = LI\,dI/dt = (0.500)(1.26)(36.8) = 23.2\ \mathrm{W}$, agreeing with the balance to the rounding.
Four parts, and only one of them needed the exponential evaluated, because the instant asked about was one time constant.
The lesson for the exam: read part (b) against part (a) before computing anything. Examiners pick instants at one, two or three time constants far more often than at arbitrary times, and recognising that turns an exponential into a remembered number.
Practice
A · concept 4 questions
1§13.4 — an ideal coil in a settled direct current circuit●●○○○
A lamp runs from a battery through a switch. Someone adds a large ideal coil in series with it and claims the lamp will be permanently dimmer, because a coil opposes current.
Given
an ideal coil, meaning one with no resistance of its own
a battery of fixed voltage and a lamp of fixed resistance
the circuit is left switched on and has long since settled
Find
(a) True or false: the steady current is smaller with the coil than without it. Give your reason in one sentence.
Hint 1/4
The word settled is doing all the work in this question. Ask what quantity has to be changing for a coil to do anything at all.
Hint 2/4
$\mathcal{E} = -L\,dI/dt$, so an ideal coil produces a voltage only while the current is changing.
Hint 3/4
In the settled state $dI/dt = 0$, so the coil's voltage is $L\times0$, and the loop rule then contains only the battery and the lamp.
Hint 4/4
False: the steady current is exactly what it was without the coil. Only the first few time constants after switching on are different.
Show solution
We answer by evaluating the coil's own voltage in the settled state rather than by arguing from impedance, because impedance is an alternating current idea and this circuit is direct current.
settled means by definition that nothing is changing, and the coil's only output is proportional to a rate of change
$$V_0 = IR \Rightarrow I = \frac{V_0}{R}$$
with the coil contributing nothing, the loop rule is the same equation it would have been without it
Answer $$\boxed{\ \text{False; the steady current is unchanged}\ }$$
Check
Limiting case check: the claim would have to hold for any inductance, including one so small that its time constant is a nanosecond, and nobody would expect a nanosecond of delay to dim a lamp forever.
2§13.7 — what is true at resonance●●●○○
A series circuit containing a resistor, a coil and a capacitor is driven by a source whose frequency can be tuned, and the frequency is set to the one that maximises the current.
Given
$R$, $L$ and $C$ in series across a source of adjustable frequency
the frequency has been tuned so that the current is at its largest
the source voltage is held fixed as the frequency is changed
Find
(a) Which one of the following is true at that frequency?
Hint 1/4
Write down what resonance means as an equation between two quantities, then test each statement against that equation rather than against a memory of the word.
Hint 2/4
At resonance $X_L = X_C$, so $Z = R$, $I = V/R$ is at its largest, $\phi = 0$, $\cos\phi = 1$ and $P = I^{2}R$ is at its largest too.
Hint 3/4
The reactances are equal but not zero, and the element voltages are $V_L = IX_L$ and $V_C = IX_C$, both computed with the largest current the circuit ever carries.
Hint 4/4
Since $X_L$ is unchanged and $I$ is at its peak, $V_L$ is at its peak too, and whenever $X_L$ exceeds $R$ it is larger than the supply.
Show solution
Each option is tested against the impedance formula rather than against a memory of what resonance means, which is what stops "the reactances are equal" from quietly becoming "the reactances are zero".
Turn the condition into consequences
$$X_L = X_C \Rightarrow Z = R,\ \ I = \frac{V}{R}\ \text{(largest)},\ \ \phi = 0$$
each consequence is read off the impedance formula rather than recalled, which is what stops equal from turning into zero
so the coil's voltage exceeding the supply is not exotic; it happens in any circuit whose reactance at resonance beats its resistance
Answer $$\boxed{\ V_L\ \text{can exceed the supply voltage}\ }$$
Check
Independent check against the worked example above, where at resonance the supply was $120\ \mathrm{V}$ and both $V_L$ and $V_C$ came to $136\ \mathrm{V}$.
3§13.6 — average power in an ideal coil●●○○○
An ideal coil is connected alone across an alternating supply and draws a current of several amperes. An electricity meter is placed in the line to see what it costs to run.
Given
an ideal coil, no resistance
connected alone across a sinusoidal supply
the current through it is not zero
Find
(a) True or false: the average power over a complete cycle is zero. Give your reason in one sentence.
Hint 1/4
Average power is not the same question as how much current flows. Ask what the instantaneous product of voltage and current does over a cycle.
Hint 2/4
In an ideal coil the voltage leads the current by a quarter of a cycle, so $p = vI \propto \sin\omega t\cos\omega t = \tfrac12\sin2\omega t$.
Hint 3/4
A sine of twice the frequency is positive for half of each cycle and negative for the other half, in equal measure, and its average over a whole cycle is zero.
Hint 4/4
True: energy flows into the field for a quarter of a cycle and back out during the next, and nothing is consumed on average.
Show solution
We average the instantaneous power over a cycle instead of quoting $P = VI\cos\phi$, because the product form shows where the zero comes from: there is no constant term to average, rather than a small one.
Independent check by the general formula: $P = V_{\rm rms}I_{\rm rms}\cos\phi$ with $\phi = 90^{\circ}$ gives zero for any voltage and any current whatever.
4§13.2 — doubling the turns of a solenoid●●○○○
A solenoid is rewound with twice as many turns of thinner wire. Its length and its cross sectional area are unchanged, and it still has air inside.
Given
the number of turns is doubled, from $N$ to $2N$
the length $\ell$ and the cross sectional area $A$ are unchanged
$L_{\rm sol} = \mu_0N^{2}A/\ell$
Find
(a) What happens to the self-inductance?
Hint 1/4
The turns do two separate jobs in this device, and the question is really asking how many of them changed.
Hint 2/4
$L = \mu_0N^{2}A/\ell$: one factor of $N$ comes from the field the winding makes, the other from the linkage of that field through the winding.
Hint 3/4
Replacing $N$ by $2N$ with $A$ and $\ell$ held fixed gives $\mu_0(2N)^{2}A/\ell = 4\mu_0N^{2}A/\ell$.
Hint 4/4
The inductance is four times what it was.
Show solution
We take the ratio instead of substituting numbers twice; $\mu_0$, $A$ and $\ell$ all cancel, so no values are needed and none can be mistyped.
Take the ratio rather than substituting twice
$$\frac{L'}{L} = \frac{(2N)^{2}}{N^{2}} = 4$$
everything except the turn count cancels, so no numerical values are needed and none can be got wrong
Answer $$\boxed{\ L' = 4L\ }$$
Check
Independent check against the numeric table in the self-inductance concept: $400$ turns gave $0.569\ \mathrm{mH}$ and $800$ turns gave $2.27\ \mathrm{mH}$, and $2.27/0.569 = 3.99$.
B · computation 7 questions
1§13.1 — mutual inductance from the geometry of two windings●●●○○
A long solenoid of length $0.400\ \mathrm{m}$ has $800$ turns and a cross sectional area of $6.00\ \mathrm{cm^{2}}$. A secondary winding of $120$ turns is wrapped tightly around its middle.
Independent check by scaling from the first worked example, which had $N_1N_2A/\ell = 40.0$ and gave $M = 5.03\times10^{-5}\ \mathrm{H}$. Here the same combination is $144$, which is $3.6$ times larger, and $3.6\times5.03\times10^{-5} = 1.81\times10^{-4}\ \mathrm{H}$.
2§13.2 — self-inductance from a measured flux, and a current reversal●●●○○
A coil of $250$ turns is carrying $1.50\ \mathrm{A}$, and a measurement shows a magnetic flux of $4.80\times10^{-5}\ \mathrm{Wb}$ through each turn. The current is then reversed steadily, from $1.50\ \mathrm{A}$ one way to $1.50\ \mathrm{A}$ the other, in $40.0\ \mathrm{ms}$.
Given
$N = 250$ turns, $I = 1.50\ \mathrm{A}$
$\Phi_B = 4.80\times10^{-5}\ \mathrm{Wb}$ through each turn at that current
the reversal takes $40.0\times10^{-3}\ \mathrm{s}$ and is steady
Find
(a) Find the self-inductance of the coil.
(b) Find the size of the voltage across it during the reversal.
Hint 1/4
The word reversed is the trap in part (b): decide how much the current actually changes before dividing by anything.
Hint 2/4
$L = N\Phi_B/I$ and $|\mathcal{E}| = L|\Delta I|/\Delta t$.
Hint 3/4
With $N = 250$, $\Phi_B = 4.80\times10^{-5}\ \mathrm{Wb}$ and $I = 1.50\ \mathrm{A}$; and the current goes from $+1.50\ \mathrm{A}$ to $-1.50\ \mathrm{A}$, a change of $3.00\ \mathrm{A}$, in $40.0\ \mathrm{ms}$.
The inductance comes from the flux linkage definition rather than from geometry, because a measured flux is handed to us and no dimensions are; the reversal is then treated as a change of $3.00\ \mathrm{A}$, which is where this question is usually halved.
the current passes through zero and out the other side, so the change is twice the starting value; treating a reversal as a switch off is the standard way to be out by two here
Independent check by units on part (a): a weber divided by an amp is a henry, and $250\times4.80\times10^{-5} = 1.20\times10^{-2}\ \mathrm{Wb}$ of linkage over $1.50\ \mathrm{A}$ gives $8.00\ \mathrm{mH}$ without any other quantity entering.
3§13.3 — stored energy checked against the energy density●●●●○
A solenoid $0.300\ \mathrm{m}$ long is wound with $500$ turns on a form of radius $2.00\ \mathrm{cm}$ and carries $2.50\ \mathrm{A}$.
Both routes are computed on purpose so that their agreement is the check; either one alone would answer the question, but neither alone would catch a slipped power of ten in the field.
the volume is the inside of the solenoid, where the field is, and nothing outside it counts
Answer $$\boxed{\ L = 1.32\ \mathrm{mH},\ U = 4.11\ \mathrm{mJ},\ B = 5.24\ \mathrm{mT},\ u = 10.9\ \mathrm{J/m^{3}}\ }$$
Check
The two routes agreeing to three figures is the verification, and it is independent: the first used $L$ and never mentioned $B$, the second used $B$ and never mentioned $L$.
4§13.4 — decay of the current in a coil and resistor loop●●●○○
A $0.400\ \mathrm{H}$ coil is carrying $1.50\ \mathrm{A}$ in a loop with a total resistance of $20.0\ \mathrm{\Omega}$ when the source is removed and the loop is closed on itself.
$I_0 = 1.50\ \mathrm{A}$ at the moment the source is removed
$I(t) = I_0e^{-t/\tau}$
Find
(a) Find the time constant and the current at $t = 30.0\ \mathrm{ms}$.
(b) Find the time at which the current has fallen to $0.500\ \mathrm{A}$.
(c) Find the total energy dissipated in the resistor by the time the current has died away.
Hint 1/4
Part (c) does not need the exponential at all. Ask where the energy in the resistor can possibly have come from.
Hint 2/4
$\tau = L/R$, $I = I_0e^{-t/\tau}$, $t = -\tau\ln(I/I_0)$, and the energy dissipated in total equals the energy $\tfrac12LI_0^{2}$ that the coil started with.
Hint 3/4
With $L = 0.400\ \mathrm{H}$ and $R = 20.0\ \mathrm{\Omega}$ the time constant is $20.0\ \mathrm{ms}$; the current asked about in (b) is one third of $1.50\ \mathrm{A}$.
Hint 4/4
$\tau = 20.0\ \mathrm{ms}$, $I(30.0\ \mathrm{ms}) = 0.335\ \mathrm{A}$, $t = \tau\ln3 = 22.0\ \mathrm{ms}$, and the energy is $0.450\ \mathrm{J}$.
Show solution
The heat is taken from the stored energy rather than by integrating $I^{2}R$ over the decay, because the source has been disconnected and the field's whole store has exactly one place it can go.
with the source disconnected, the only energy available is what the field held, and the only place it can go is the resistor
Answer $$\boxed{\ \tau = 20.0\ \mathrm{ms},\ I = 0.335\ \mathrm{A},\ t = 22.0\ \mathrm{ms},\ U = 0.450\ \mathrm{J}\ }$$
Check
Independent check on part (c) by integrating instead: $\int_0^{\infty}I_0^{2}Re^{-2t/\tau}dt = I_0^{2}R\tau/2 = (2.25)(20.0)(0.0200)/2 = 0.450\ \mathrm{J}$, agreeing with the energy argument.
5§13.5 — a charged capacitor released into a coil●●●○○
A $2.50\ \mathrm{\mu F}$ capacitor is charged to $12.0\ \mathrm{V}$ and then connected across a $40.0\ \mathrm{mH}$ coil of negligible resistance.
Given
$C = 2.50\times10^{-6}\ \mathrm{F}$ charged to $V_0 = 12.0\ \mathrm{V}$
$L = 40.0\times10^{-3}\ \mathrm{H}$, no resistance
the current is zero at the moment of connection
Find
(a) Find the angular frequency and the frequency in hertz.
(b) Find the peak charge and the peak current.
(c) Find the total energy, by two different routes.
Hint 1/4
The peak charge is available before any oscillation is considered: it is simply what the capacitor was charged to.
With $L = 40.0\ \mathrm{mH}$ and $C = 2.50\ \mathrm{\mu F}$ the product is $LC = 1.00\times10^{-7}\ \mathrm{s^{2}}$, and the capacitor was charged to $12.0\ \mathrm{V}$.
The peak charge is fixed by the charging voltage before the coil is ever connected, so $Q_0 = CV_0$ comes first and the current follows from $I_0 = \omega_0Q_0$; going through an energy balance would give the same current after two more lines.
Independent check on the peak current by energy alone, bypassing the frequency: $I_0 = \sqrt{2U/L} = \sqrt{2(1.80\times10^{-4})/(4.00\times10^{-2})} = 9.49\times10^{-2}\ \mathrm{A}$.
6§13.6 — finding the frequency that gives a stated reactance●●●●○
A $25.0\ \mathrm{mH}$ coil and a $0.500\ \mathrm{\mu F}$ capacitor are available, and a signal generator of adjustable frequency is used to test them one at a time.
Given
$L = 25.0\times10^{-3}\ \mathrm{H}$
$C = 0.500\times10^{-6}\ \mathrm{F}$
$X_L = 2\pi fL$ and $X_C = 1/(2\pi fC)$
Find
(a) At what frequency does the coil have a reactance of $400\ \mathrm{\Omega}$?
(b) At what frequency does the capacitor have a reactance of $400\ \mathrm{\Omega}$?
(c) At what frequency are the two reactances equal, and what is their common value there?
Hint 1/4
The first two parts are the same formula rearranged for $f$; the third asks where two curves cross and needs no target value at all.
Hint 2/4
$f = X_L/(2\pi L)$, $f = 1/(2\pi CX_C)$, and $X_L = X_C$ at $f_0 = 1/(2\pi\sqrt{LC})$ where both equal $\sqrt{L/C}$.
Hint 3/4
With $L = 25.0\ \mathrm{mH}$, $C = 0.500\ \mathrm{\mu F}$ and a target reactance of $400\ \mathrm{\Omega}$ in the first two parts.
Hint 4/4
$2.55\ \mathrm{kHz}$, $796\ \mathrm{Hz}$, and the crossing at $1.42\ \mathrm{kHz}$ where both are $224\ \mathrm{\Omega}$.
Show solution
Each reactance formula is rearranged for the frequency rather than solved by trial, and the shared value at the crossing is then taken from $X_LX_C = L/C$, which is free of frequency and so needs no $f_0$ at all.
Independent check on part (c) by substituting into the inductive formula, which was not used to obtain it: $X_L = 2\pi(1.42\times10^{3})(25.0\times10^{-3}) = 224\ \mathrm{\Omega}$.
7§13.7 — a full series solve with the element voltages●●●●○
A $150\ \mathrm{\Omega}$ resistor, a $0.250\ \mathrm{H}$ coil and a $8.00\ \mathrm{\mu F}$ capacitor are in series across a $90.0\ \mathrm{V}$ root mean square supply at $100\ \mathrm{Hz}$.
Given
$R = 150\ \mathrm{\Omega}$, $L = 0.250\ \mathrm{H}$, $C = 8.00\times10^{-6}\ \mathrm{F}$, in series
$V_{\rm rms} = 90.0\ \mathrm{V}$ at $f = 100\ \mathrm{Hz}$
the coil has no resistance of its own
Find
(a) Find the impedance and the root mean square current.
(b) Find the phase angle and the average power.
(c) Find the root mean square voltage across each of the three elements, and check that they recombine to the supply.
Hint 1/4
One current runs through all three elements, so once it is known, part (c) is three multiplications.
With $f = 100\ \mathrm{Hz}$, $\omega = 628\ \mathrm{rad/s}$; then $X_L = 157\ \mathrm{\Omega}$ and $X_C = 199\ \mathrm{\Omega}$, against $R = 150\ \mathrm{\Omega}$ and a supply of $90.0\ \mathrm{V}$.
Straight down the method box, and the element voltages are checked by recombining them at right angles rather than by adding them, because their plain sum exceeds the supply and looks alarming for no reason.
the reactive readings are subtracted from each other first, and only then combined with the resistive one at right angles
Answer $$\boxed{\ Z = 156\ \mathrm{\Omega},\ I = 0.578\ \mathrm{A},\ \phi = -15.6^{\circ},\ P = 50.1\ \mathrm{W};\ V_R = 86.7,\ V_L = 90.8,\ V_C = 115\ \mathrm{V}\ }$$
Check
Independent check on the power from the other formula: $\cos(-15.6^{\circ}) = 0.963$, so $V_{\rm rms}I_{\rm rms}\cos\phi = (90.0)(0.578)(0.963) = 50.1\ \mathrm{W}$, matching without using the resistance.
C · exam level 3 questions
1§13.4 — a coil with its own winding resistance●●●●○
A coil of self-inductance $0.800\ \mathrm{H}$ has a winding resistance of $12.0\ \mathrm{\Omega}$. It is connected in series with a $28.0\ \mathrm{\Omega}$ resistor across a $24.0\ \mathrm{V}$ source of negligible internal resistance, and the switch is closed at $t = 0$ with no current flowing.
Given
coil: $L = 0.800\ \mathrm{H}$ with a winding resistance of $12.0\ \mathrm{\Omega}$
external resistor: $28.0\ \mathrm{\Omega}$
$V_0 = 24.0\ \mathrm{V}$, current zero at $t = 0$
$I(t) = (V_0/R_{\rm tot})(1-e^{-t/\tau})$ with $\tau = L/R_{\rm tot}$
Find
(a) Find the time constant and the final current.
(b) Find the current at $t = 10.0\ \mathrm{ms}$.
(c) Find the energy eventually stored in the coil's field.
(d) In the final state, find the total power dissipated and how much of it is dissipated inside the coil itself.
Hint 1/4
The coil is two components drawn as one. Separate them on paper before writing any formula: an ideal inductance and a resistor in series.
Hint 2/4
$R_{\rm tot}$ is the sum of every resistance in the loop, and both $\tau = L/R_{\rm tot}$ and $I_{\max} = V_0/R_{\rm tot}$ use that total.
Hint 3/4
Here $R_{\rm tot} = 12.0+28.0 = 40.0\ \mathrm{\Omega}$ with $L = 0.800\ \mathrm{H}$ and $V_0 = 24.0\ \mathrm{V}$, and the instant asked about is half a time constant.
Hint 4/4
$\tau = 20.0\ \mathrm{ms}$, $I_{\max} = 0.600\ \mathrm{A}$, $I(10.0\ \mathrm{ms}) = 0.236\ \mathrm{A}$, $U = 0.144\ \mathrm{J}$, and the powers are $14.4\ \mathrm{W}$ in total with $4.32\ \mathrm{W}$ inside the coil.
Show solution
The winding resistance is folded into one total before any formula is written; leaving it out lengthens the time constant and raises the final current, and both errors push the same way, so nothing in the answer looks wrong afterwards.
Independent check on the total power from the source side rather than the resistor side: $P = V_0I = (24.0)(0.600) = 14.4\ \mathrm{W}$, which must match once the coil has stopped absorbing energy.
2§13.7 — the capacitor voltage at resonance●●●●○
A series circuit of $R = 20.0\ \mathrm{\Omega}$, $L = 5.00\ \mathrm{mH}$ and $C = 2.00\ \mathrm{\mu F}$ is driven at exactly its resonant frequency by a $10.0\ \mathrm{V}$ root mean square source.
Given
$R = 20.0\ \mathrm{\Omega}$, $L = 5.00\times10^{-3}\ \mathrm{H}$, $C = 2.00\times10^{-6}\ \mathrm{F}$, in series
$V_{\rm rms} = 10.0\ \mathrm{V}$
the driving frequency is the resonant one
Find
(a) What root mean square voltage would a meter read across the capacitor?
Hint 1/4
Two steps hide in this question: what the current is at resonance, and what the capacitor does with that current.
Hint 2/4
At resonance $Z = R$, so $I = V/R$; and $V_C = IX_C$ with $X_C = 1/(\omega_0C)$ and $\omega_0 = 1/\sqrt{LC}$.
Hint 3/4
With $L = 5.00\ \mathrm{mH}$ and $C = 2.00\ \mathrm{\mu F}$, $\omega_0 = 1.00\times10^{4}\ \mathrm{rad/s}$ and $X_C = 50.0\ \mathrm{\Omega}$, while $I = 10.0/20.0$.
Hint 4/4
$I = 0.500\ \mathrm{A}$ and $V_C = (0.500)(50.0) = 25.0\ \mathrm{V}$, two and a half times the supply.
Show solution
The current is obtained from $Z = R$ first and the capacitor voltage from $IX_C$ afterwards, rather than reasoning about the voltage directly, because the cancellation happens round the loop and not inside the component.
the reactance did not cancel inside the capacitor; it cancelled only in the sum round the loop
Answer $$\boxed{\ V_C = 25.0\ \mathrm{V}\ }$$
Check
Independent check by the ratio $X_L/R = 50.0/20.0 = 2.50$, which is the factor by which the reactive voltages exceed the supply at resonance, and $2.50\times10.0\ \mathrm{V} = 25.0\ \mathrm{V}$.
3§13.5 — six lines of an oscillation, two of them wrong●●●●○
A $20.0\ \mathrm{mH}$ coil of negligible resistance is connected across a $5.00\ \mathrm{\mu F}$ capacitor that has been charged to $8.00\ \mathrm{V}$. A student is asked for the frequency, the peak current and the total energy, and produces the six lines below, reporting $1.59\ \mathrm{MHz}$, $400\ \mathrm{A}$ and $2.00\times10^{-5}\ \mathrm{J}$. Some of the lines are faulty; you are not told how many.
line 1: $LC = (2.00\times10^{-2})(5.00\times10^{-6}) = 1.00\times10^{-7}\ \mathrm{s^{2}}$
line 2: $\omega_0 = 1/(LC) = 1.00\times10^{7}\ \mathrm{rad/s}$
line 3: $f = \omega_0/2\pi = 1.59\times10^{6}\ \mathrm{Hz}$
line 4: $Q_0 = CV_0 = 4.00\times10^{-5}\ \mathrm{C}$
line 5: $I_0 = \omega_0Q_0 = 400\ \mathrm{A}$
line 6: $U = \tfrac12CV_0 = 2.00\times10^{-5}\ \mathrm{J}$
Find
(a) Which lines are faulty?
(b) Give the corrected frequency, peak current and total energy.
Hint 1/4
Before hunting, apply a plausibility test to the reported answers. A four hundred amp current out of a capacitor charged to eight volts should not survive a second glance.
Hint 2/4
$\omega_0 = 1/\sqrt{LC}$, not $1/(LC)$, and $U = \tfrac12CV_0^{2}$, with the voltage squared.
Hint 3/4
With $LC = 1.00\times10^{-7}\ \mathrm{s^{2}}$ the root is $3.16\times10^{-4}\ \mathrm{s}$; and with $C = 5.00\ \mathrm{\mu F}$ and $V_0 = 8.00\ \mathrm{V}$ the energy needs $V_0^{2} = 64.0\ \mathrm{V^{2}}$.
Hint 4/4
Lines 2 and 6 are the faulty ones, and the corrected answers are $f = 503\ \mathrm{Hz}$, $I_0 = 0.126\ \mathrm{A}$ and $U = 1.60\times10^{-4}\ \mathrm{J}$.
Show solution
We test the reported numbers for plausibility before hunting through the algebra: one implausible pair pins the faulty line in a single step, while checking six lines in order costs six.
forty microcoulombs delivered at four hundred amps would be over in a tenth of a microsecond, which contradicts the reported frequency and flags the frequency line rather than the current line
the voltage was not squared, which cost a factor of eight and, unlike the first fault, produces an answer that looks entirely reasonable
Answer $$\boxed{\ \text{lines 2 and 6};\quad f = 503\ \mathrm{Hz},\ I_0 = 0.126\ \mathrm{A},\ U = 1.60\times10^{-4}\ \mathrm{J}\ }$$
Check
Independent check that the repairs are consistent with each other: $\tfrac12LI_0^{2} = \tfrac12(2.00\times10^{-2})(0.126)^{2} = 1.60\times10^{-4}\ \mathrm{J}$, the same energy as the capacitor route, which ties the corrected frequency and the corrected energy together.
The dangerous fault was the second one. The first produced an absurd four hundred amps and would have been caught by anyone glancing at the answer; the second produced twenty microjoules where the truth is a hundred and sixty, and nothing about it looks wrong.
D · interleaved 3 questions
1§13.5 — one capacitor, two circuits●●●●○
A $100\ \mathrm{\mu F}$ capacitor is charged from a $20.0\ \mathrm{V}$ source through a $2.00\ \mathrm{k\Omega}$ resistor, starting empty. Once it is fully charged it is disconnected from that circuit and connected instead across a $50.0\ \mathrm{mH}$ coil of negligible resistance.
charging: $V_0 = 20.0\ \mathrm{V}$ through $R = 2.00\times10^{3}\ \mathrm{\Omega}$
then: the charged capacitor alone with $L = 50.0\times10^{-3}\ \mathrm{H}$
$q(t) = Q_f(1-e^{-t/RC})$ for the charging stage
Find
(a) Find the charging time constant and the time to reach ninety per cent of full charge.
(b) Find the energy stored in the capacitor when it is fully charged.
(c) Find the frequency of the oscillation once the coil is connected.
(d) Find the peak current in that oscillation, and check it against the energy from part (b).
Hint 1/4
The capacitor is the only thing the two halves share. The first half is a question from the direct current work and the second is a question from this week; the link between them is the stored energy.
Hint 2/4
$\tau = RC$ and $t = -\tau\ln(1-q/Q_f)$; then $U = \tfrac12CV_0^{2}$, $\omega_0 = 1/\sqrt{LC}$, $Q_0 = CV_0$ and $I_0 = \omega_0Q_0$.
Hint 3/4
With $R = 2.00\ \mathrm{k\Omega}$ and $C = 100\ \mathrm{\mu F}$ the charging time constant is $0.200\ \mathrm{s}$; the capacitor ends at $20.0\ \mathrm{V}$; and the coil is $50.0\ \mathrm{mH}$.
Hint 4/4
$\tau = 0.200\ \mathrm{s}$ and $t = 0.461\ \mathrm{s}$; $U = 0.0200\ \mathrm{J}$; $f = 71.2\ \mathrm{Hz}$; $I_0 = 0.894\ \mathrm{A}$, which returns $0.0200\ \mathrm{J}$ through $\tfrac12LI_0^{2}$.
Show solution
The two stages are kept apart and the only thing carried across is the charge on the capacitor; leaving the resistor in the second stage is the tempting error and would put a decay where the physics has an oscillation.
nearly an amp from a capacitor that took half a second to charge at a tenth of that current, which is what a resonant circuit does to a rate
Answer $$\boxed{\ \tau = 0.200\ \mathrm{s},\ t = 0.461\ \mathrm{s},\ U = 0.0200\ \mathrm{J},\ f = 71.2\ \mathrm{Hz},\ I_0 = 0.894\ \mathrm{A}\ }$$
Check
Independent check on part (d) against part (b): $\tfrac12LI_0^{2} = \tfrac12(5.00\times10^{-2})(0.894)^{2} = 2.00\times10^{-2}\ \mathrm{J}$, the same energy, reached through the coil rather than the capacitor.
The charging current never exceeded $V_0/R = 10.0\ \mathrm{mA}$, yet the oscillation peaks at nearly ninety times that. Energy, not current, is what carried across.
2§13.2 — from a winding to a field to a voltage●●●●○
A solenoid $0.250\ \mathrm{m}$ long is wound with $600$ turns on a form of radius $1.20\ \mathrm{cm}$ and carries $4.00\ \mathrm{A}$. The current is then switched off steadily over $5.00\ \mathrm{ms}$.
the current falls steadily to zero in $5.00\times10^{-3}\ \mathrm{s}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the field well inside the solenoid.
(b) Find the magnetic flux through one turn.
(c) Find the self-inductance from the flux linkage, and check it against the geometric formula.
(d) Find the size of the voltage across the coil during the switch off.
Hint 1/4
This is one chain with four links, and each part is the input to the next. The only new step is the last one.
Hint 2/4
$B = \mu_0nI$ with $n = N/\ell$; $\Phi_B = BA$; $L = N\Phi_B/I$, which should equal $\mu_0N^{2}A/\ell$; and $|\mathcal{E}| = L|\Delta I|/\Delta t$.
Hint 3/4
With $N = 600$ over $\ell = 0.250\ \mathrm{m}$ giving $n = 2400\ \mathrm{m^{-1}}$, an area $A = \pi(1.20\times10^{-2})^{2} = 4.52\times10^{-4}\ \mathrm{m^{2}}$, a current of $4.00\ \mathrm{A}$ and a switch off time of $5.00\ \mathrm{ms}$.
The chain is walked once from the winding through to the voltage, and the inductance is then got a second way from the geometry, because two agreeing values are the cheapest check available on a five step calculation.
turns per metre before the field, as always, and twelve millitesla is a couple of hundred times the Earth's field, which is right for a coil of this size
steadily means the rate is a single number, so no exponential enters this part
Answer $$\boxed{\ B = 1.21\times10^{-2}\ \mathrm{T},\ \Phi_B = 5.46\times10^{-6}\ \mathrm{Wb},\ L = 8.19\times10^{-4}\ \mathrm{H},\ |\mathcal{E}| = 0.655\ \mathrm{V}\ }$$
Check
Independent check on part (d) through energy rather than through the inductance: the coil held $\tfrac12LI^{2} = 6.55\ \mathrm{mJ}$, and delivering that in $5.00\ \mathrm{ms}$ at an average current of $2.00\ \mathrm{A}$ needs an average voltage of $6.55\ \mathrm{mJ}/(2.00\ \mathrm{A}\times5.00\ \mathrm{ms}) = 0.655\ \mathrm{V}$.
3§13.6 — a heater with and without a coil in series●●●●○
A heating element of resistance $40.0\ \mathrm{\Omega}$ is run from a $220\ \mathrm{V}$ root mean square supply at $50.0\ \mathrm{Hz}$. Someone then puts a $0.150\ \mathrm{H}$ coil of negligible resistance in series with it, to cut the heat without wasting energy in a series resistor.
Given
heating element: $R = 40.0\ \mathrm{\Omega}$
supply: $220\ \mathrm{V}$ root mean square at $50.0\ \mathrm{Hz}$
added coil: $L = 0.150\ \mathrm{H}$, no resistance of its own
Find
(a) Find the average power in the element on its own.
(b) Find the current and the average power once the coil is in series.
(c) What steady direct voltage across the element alone would produce the same heating as in part (a)?
Hint 1/4
Part (c) is not asking you to compute anything new; it is asking what the root mean square value of an alternating voltage was defined to mean.
Hint 2/4
$P = V_{\rm rms}^{2}/R$ for a resistor alone; with a coil in series, $Z = \sqrt{R^{2}+X_L^{2}}$, $I = V/Z$ and $P = I^{2}R$.
Hint 3/4
With $R = 40.0\ \mathrm{\Omega}$ on $220\ \mathrm{V}$; and with $X_L = 2\pi(50.0)(0.150) = 47.1\ \mathrm{\Omega}$ added in series.
Hint 4/4
$P = 1210\ \mathrm{W}$; then $Z = 61.8\ \mathrm{\Omega}$, $I = 3.56\ \mathrm{A}$ and $P = 507\ \mathrm{W}$; and the equivalent direct voltage is $220\ \mathrm{V}$.
Show solution
The power is taken as $I^{2}R$ and not as $V_{\rm rms}^{2}/Z$ or $I^{2}Z$: the impedance route would claim $783\ \mathrm{W}$ and destroy the whole point of putting the coil there.
the resistance and not the impedance, which matters more here than usual: using $Z$ would have claimed $783\ \mathrm{W}$ and hidden the whole point of the scheme
Independent check on part (b) by the power factor route: $\cos\phi = R/Z = 40.0/61.8 = 0.647$, so $P = V_{\rm rms}I_{\rm rms}\cos\phi = (220)(3.56)(0.647) = 507\ \mathrm{W}$.
A series resistor of $21.8\ \mathrm{\Omega}$ would have cut the element to exactly the same $507\ \mathrm{W}$, since it takes a total of $61.8\ \mathrm{\Omega}$ to hold the current at $3.56\ \mathrm{A}$, but that resistor would itself have burned a further $276\ \mathrm{W}$. The coil achieves the same reduction and burns nothing, which is why chokes rather than resistors are used to limit alternating currents.
Mistake ledger (23 entries)
⚠ Treating the mutual inductance as belonging to one coil
the symbol is written next to whichever coil is being driven, so it starts to look like a property of that coil in the way resistance is a property of a resistor
wrong$$M_{12} \neq M_{21}$$
right$$M_{12} = M_{21} = M$$
⚠ Using the area of the wrong coil
the flux being counted belongs to coil 2, so its own area feels like the right one to reach for
wrong$$\Phi_{21} = B_1A_2$$
right$$\Phi_{21} = B_1A_{\rm solenoid}$$
⚠ Multiplying by the current instead of by its rate of change
the definition contains a current and the formula for the voltage does not, and the two get merged into one half remembered line
wrong$$\mathcal{E}_2 = -MI_1$$
right$$\mathcal{E}_2 = -M\dfrac{dI_1}{dt}$$
⚠ Putting the total turns where turns per metre belong
the two forms of the solenoid formula look almost identical and both are correct, so the wrong one gets copied from memory under time pressure
power uses the resistance and never the impedance; at resonance $Z = R$, $\phi = 0$ and the current peaks
Check yourself
Close the page and write out from memory: the definition of mutual inductance and the voltage it produces; the definition of self-inductance, the solenoid formula for it, and why the turn count is squared; the two expressions for the energy stored in a coil and where that energy sits; the growth and decay laws of an LR circuit, its time constant, and what happens at the instant a switch is opened; the natural frequency of an LC pair, how the energy divides between the two components, and what a resistance does; the reactance of a coil and of a capacitor and their opposite dependence on frequency, together with the average power in each; and the impedance, phase angle, power and resonance condition of a driven series circuit.
Compute the mutual inductance of a coil wound over a solenoid from the geometry alone, and use a measurement made in one direction to answer a question about the other?
c-mutual-inductance
Get the self-inductance of a coil from either its geometry or a measured flux linkage, and decide in advance whether the formula in front of you wants $N$ or $n$?
c-self-inductance
Produce the stored energy from the inductance and the current, then reproduce the same number from the field and the volume, without looking up either formula?
c-energy-density
State the current at the first instant and long afterwards for a coil switched on, extract a time from the exponential by isolating it before taking a logarithm, and explain where the spark at an opening switch gets its voltage?
c-lr-circuit
Write the natural frequency without losing the square root, find the peak current from the peak charge, and say at what fraction of the peak charge the energy is evenly divided?
c-lc-oscillation
Convert a supply frequency into radians per second, get both reactances, and say without hesitating which one grows with frequency and why neither consumes power?
c-ac-reactance
Run the six lines of a series circuit in order, get the sign of the phase angle right from the reactances alone, use $R$ rather than $Z$ in the power, and explain how a component voltage can exceed the supply?
c-lrc-series
Glossary (25 terms)
mutual inductancekarşılıklı indüktans
The flux linkage produced in one coil per unit of current in another, in henrys. It belongs to the pair of coils rather than to either one, and it has the same value whichever coil is treated as the driver.
self-inductanceöz indüktans
The flux linkage of a coil per unit of its own current, in henrys. It is fixed by the geometry of the winding and does not depend on the current flowing.
henryhenry
The unit of inductance. A coil has one henry if a current changing at one ampere per second produces one volt across it. Equivalently a weber of flux linkage per ampere, or an ohm second.
inductorbobin
A circuit element whose purpose is its inductance, usually a coil of wire. An ideal one has no resistance, so a real coil is treated as an ideal inductance in series with a resistor.
ters elektromotor kuvvet
The voltage a coil produces across itself when its own current changes, always in the sense that opposes the change. It is the reason a current cannot start or stop instantly in a coil.
flux linkageakı halkalanması
The magnetic flux through one turn multiplied by the number of turns, in webers. It is the quantity that actually appears in Faraday's law for a coil, and the quantity inductance is defined from.
indüktif zaman sabiti
The quantity $L/R$ for a coil and resistor loop, in seconds. The current covers sixty three per cent of its journey in one of them, and unlike the capacitive case a larger resistance makes the circuit faster.
magnetic energy densitymanyetik enerji yoğunluğu
The energy stored per unit volume in a magnetic field, $B^{2}/2\mu_0$, in joules per cubic metre. Multiplying it by the volume the field occupies gives the same total as the inductance formula.
LC circuitLC devresi
A loop containing only a capacitor and a coil. Its charge and current oscillate freely, with energy passing between the electric field of the capacitor and the magnetic field of the coil.
The periodic exchange of energy between a capacitor and a coil in a closed loop, obeying the same equation as a mass on a spring with the charge playing the part of the displacement.
natural angular frequencydoğal açısal frekans
The angular frequency $1/\sqrt{LC}$ at which an undriven capacitor and coil oscillate, in radians per second. It depends on the components alone and not on how much charge was put in.
sönümlü salınım
An oscillation in a loop that also contains resistance, so that the amplitude decreases with every cycle and the frequency is slightly below the undamped value.
kritik sönüm
The condition $R = 2\sqrt{L/C}$ at which a circuit stops oscillating altogether. Above it the charge falls away without ever crossing zero.
alternating currentalternatif akım
A current that reverses direction periodically, here always sinusoidally. The circuit quantities are quoted either as peak values or as root mean square values, and the two differ by a factor of the square root of two.
root mean square valueetkin değer
The steady value that would deliver the same average heating in a resistor as the alternating one does. For a sine wave it is the peak divided by the square root of two, and it is what meters and quoted supply voltages give.
reactancereaktans
The ratio of voltage to current for a coil or a capacitor in an alternating circuit, in ohms. It limits the current exactly as a resistance would but consumes no average power, because the voltage and the current are a quarter of a cycle apart.
inductive reactanceindüktif reaktans
The reactance $\omega L$ of a coil, which grows in proportion to frequency, so that a coil obstructs rapid changes and passes slow ones.
capacitive reactancekapasitif reaktans
The reactance $1/(\omega C)$ of a capacitor, which falls as the frequency rises, so that a capacitor passes rapid changes and blocks slow ones.
impedanceempedans
The single number in ohms relating the source voltage to the current in an alternating circuit. In a series loop it is the resistance combined at right angles with the difference of the two reactances, never their sum.
phase anglefaz açısı
The angle by which the source voltage leads the current in a driven circuit. It is positive when the inductive reactance dominates and negative when the capacitive one does.
fazör
An arrow whose length is the peak value of a sinusoidal quantity and whose direction records its timing relative to a chosen reference. Adding phasors as vectors is what replaces adding voltages arithmetically.
resonancerezonans
The condition in a driven series circuit where the two reactances are equal and cancel. The impedance falls to the resistance alone, the current and the power reach their largest values, and the phase angle is zero.
resonant frequencyrezonans frekansı
The driving frequency at which resonance occurs, equal to the natural frequency of the same coil and capacitor. Below it a series circuit behaves capacitively and above it inductively.
power factorgüç katsayısı
The cosine of the phase angle, equal to the resistance divided by the impedance. It is the fraction of the apparent power that is actually consumed, and it is one at resonance and zero for a purely reactive circuit.
chokeşok bobini
A coil used deliberately to limit an alternating current. It does the job a series resistor would do but without dissipating power, since its average power consumption is zero.
What comes next
§14 · Maxwell's Equations and Electromagnetic Waves
Every circuit on this page kept its energy inside a loop of wire, and the fastest of them was oscillating at a few hundred hertz.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook, whose treatment of inductance, electromagnetic oscillations and alternating current covers the same ground as this section. Its end of chapter problems on transformers and on parallel resonant circuits run beyond what is treated here.
Course syllabus, week 13 line The scope of this section comes from the week line, which reads Inductance, Electromagnetic Oscillations, and AC Circuits, and from nowhere else. The line carries no chapter or section number, so none is quoted on this page.
Course syllabus, catalogue description and assessment weights The catalogue description lists inductance among the topics of the course, which is what places this material inside it, and the assessment weights are the only basis for anything said here about examinations.
SI definitions and constants The permeability of free space is taken as $4\pi\times10^{-7}\ \mathrm{T\,m/A}$ and the permittivity of free space as $8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$ throughout, both to well beyond the three figures used here. The henry is the derived unit of inductance, one weber per ampere.