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08Direct-current circuits: real batteries, networks, Kirchhoff's rules, and the RC circuit
Turn the key on a cold morning and the headlights dim for a second, then come back to full brightness. Nobody touched the bulbs, and the wire to them did not change. Something in the box under the bonnet decided, on its own, to hand out fewer volts for that second, and everything you know so far says a source labelled twelve volts hands out twelve volts.
By the end of this section you can take any arrangement of batteries, resistors and one capacitor, and produce the current in every , the potential difference across every element, the power delivered and dissipated everywhere, and, when a switch is thrown, the whole time course of the charge on the capacitor.
In 60 seconds
Every source has an internal resistance, so what it delivers depends on what you ask of it; networks that reduce are handled by two rules about what is shared, networks that do not reduce are handled by charge conservation at junctions and by the fact that a round trip returns you to the same potential, and a capacitor in the circuit turns all of it into an exponential in time.
of a real source
$$V_{ab} = \varepsilon - Ir$$
current I is being drawn out of the source; the sign in front of Ir flips to plus when current is driven backwards into it
whenever a reading is quoted and you are asked whether it is the true value
Three most common mistakes
Treating the number printed on the battery as the voltage across the external circuit. It is the voltage across the external circuit only when no current flows; the moment current is drawn, the internal resistance takes its cut.
Adding resistors in parallel, or adding the reciprocals and forgetting to invert at the end. A parallel combination is always smaller than the smallest resistor in it, and that one sentence catches both errors.
Deciding that two elements are in series or in parallel by how the picture is drawn. Series means the same current with no junction in between; parallel means both ends joined to the same two conductors. Nothing else counts.
The published assessment weights for this course are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent in total, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. No finer breakdown than that is published, so nothing is claimed here about how many marks this particular material carries.
How much time do you have?
10 minutes
You leave able to do the two things that open almost every circuit question: reduce a network of resistors to one number, and get the current out of a real battery rather than an idealised one.
The 60 second card · Formula card · Sources of emf, internal resistance, and the voltage you actually get · Resistors in series and in parallel: what is shared decides everything · Mistake ledger
45 minutes
You add the three things that separate a set-up mark from full marks: expanding a reduced network back out to every branch, writing Kirchhoff equations with the right signs, and reading an at its two ends and in between.
The 60 second card · Sources of emf, internal resistance, and the voltage you actually get · Resistors in series and in parallel: what is shared decides everything · Reduce, then expand: the whole circuit from one equivalent resistance · Kirchhoff's two rules, and the four signs you have to get right · The capacitor that fills through a resistor · Method box: reducing a network and coming back out · Method box: Kirchhoff without sign errors · Exam level example · Practice set C · Mistake ledger
full read
You can solve any two-loop network with two sources, account for every watt in it, follow a capacitor through a complete charge and discharge, say where half the battery's energy went while it charged, and decide whether a quoted meter reading is the true value or an artefact of the meter.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the terminal voltage, the current and the power of a real source with internal resistance, and extract the emf and the internal resistance from two measurements under different loads.
Classify any pair of elements as being in series, in parallel, or neither, and combine resistors accordingly into a single equivalent resistance.
Solve a reducible network completely, working inwards to the equivalent resistance and back outwards to the current, the voltage and the power in every individual element, and check the answer with the power balance.
Apply the junction rule and the loop rule with correct signs to a circuit that no amount of series and parallel combination will reduce, and interpret a negative current correctly.
Derive and use the charging law of an RC circuit, identify the time constant, and read off the current and the charge at any time or the time at which a given fraction is reached.
Analyse a discharging RC circuit, use the and the time constant interchangeably, and account for the energy stored, delivered and dissipated over a full charge and discharge cycle.
Estimate the error a real ammeter or voltmeter introduces into the circuit it is measuring, and state what resistance each meter would need in order not to disturb the reading.
Syllabus coverage
Direct-Current Circuits
Sources of emf, internal resistance and terminal voltage; resistors in series and in parallel; the complete solution of a reducible network including the power delivered and dissipated; Kirchhoff's junction and loop rules for networks that do not reduce; sources in series and in parallel and the charging of one source by another; circuits containing a resistor and a capacitor, both charging and discharging, and the steady state of a capacitor branch; ammeters and voltmeters and the disturbance a real meter causes.
The week line names the topic and carries no chapter numbers, so no chapter or section number is quoted anywhere on this page.
covered
The half life of a discharging capacitor
The time for the charge to fall to one half rather than to one over e, and the conversion between that time and the time constant.
Not named on the week line. It is kept because laboratory work quotes half lives far more often than time constants, and because converting between the two is one line. It is flagged so that nobody treats it as separately examinable on the strength of this page alone.
off_syllabus
Electrical safety and the current a human body passes
Why a given voltage is dangerous in one situation and harmless in another, and what the resistance of a body has to do with it.
Deferred. It is an application of exactly the same series circuit reasoning used here, with no new physics in it, and this page already carries a full week of new results. Nothing later in these notes depends on it.
deferred
Networks that reduce only by symmetry or by a source transformation
Balanced and unbalanced bridge networks, and the shortcut methods used to collapse them.
Deferred. The unbalanced bridge is solvable here with the junction and loop rules alone, and it is treated that way in the practice set; the shortcut methods that avoid the simultaneous equations are a separate toolkit and are not needed for anything on this page.
deferred
Recall first
Ohm's law
For an ohmic conductor the current through it and the potential difference across it are proportional: $V = IR$, with the resistance $R$ a property of the object and not of the current through it.
It is the only relation used to convert between the two unknowns in every branch of every circuit on this page.
Resistance from the geometry of a conductor
A uniform wire of length $L$, cross-sectional area $A$ and resistivity $\rho$ has $R = \rho L/A$.
Several questions here hand you a wire rather than a resistance, and this is the one line that turns one into the other.
Electrical power
A device carrying current $I$ with potential difference $V$ across it converts energy at the rate $P = IV$. For a resistance this is also $I^{2}R$ and $V^{2}/R$.
Every energy accounting on this page, and the whole of the power balance check, is built from it.
Current as a rate of flow of charge
$I = dq/dt$: the current at a point is the rate at which charge passes it.
It is what turns the loop equation of an RC circuit from a statement about voltages into a differential equation in time.
Capacitance
A capacitor holds charge $q = CV$ on either plate, where the capacitance $C$ is fixed by the geometry and the material in the gap, and it stores energy $U = q^{2}/2C = \tfrac12 CV^{2}$.
The RC blocks need both relations: the first to close the loop equation and the second to account for the energy at the end.
Conservation of charge
Charge is neither created nor destroyed; in a steady state it cannot accumulate at a point in a wire either, so whatever flows into a junction must flow out of it.
It is the entire content of the junction rule; the rule is not an extra assumption about circuits.
Potential difference is path independent
The electrostatic potential is a single valued function of position, so the work done per coulomb between two points does not depend on the route taken, and around any closed path the total change is zero.
It is the entire content of the loop rule, which is why the loop rule holds for any loop you care to draw, including ones with no source in them.
Try it yourself first (3 questions)
1§08.0 — Ohm's law and power, straight from the previous section●○○○○
Three questions before we start, to find out which tools are already sharp. Getting one wrong costs nothing; it just tells you which recall above to read twice. An electric heater is designed for a supply of $230\ \mathrm{V}$ and its element has a resistance of $46.0\ \Omega$.
Given
supply potential difference $V = 230\ \mathrm{V}$
element resistance $R = 46.0\ \Omega$, constant
the element is the only thing in the circuit
Find
(a) Find the current in the element.
(b) Find the power it converts into heat.
Hint 1/4
Two quantities are given and two are wanted, and there is one relation connecting the first pair and one connecting the second. Nothing here needs a circuit diagram.
Hint 2/4
Ohm's law is $V = IR$, and the power converted by any device is $P = IV$, which for a resistance is also $V^{2}/R$.
Hint 3/4
Substituting the given values: $V = 230\ \mathrm{V}$ and $R = 46.0\ \Omega$, with the element alone across the supply.
Hint 4/4
The current is $5.00\ \mathrm{A}$ and the power is $1.15\ \mathrm{kW}$.
the element is alone across the supply, so the whole supply voltage sits across it and no division of voltage is needed
Power, by the route that uses the answer just found
$$P = IV = (5.00)(230) = 1.15\times10^{3}\ \mathrm{W}$$
using the current just computed exposes an arithmetic slip in it, whereas going straight to $V^{2}/R$ would hide one
Answer $$\boxed{\,I = 5.00\ \mathrm{A},\qquad P = 1.15\ \mathrm{kW}\,}$$
Check
Independent route to the power, not using the current: $P = V^{2}/R = 52900/46.0 = 1.15\times10^{3}\ \mathrm{W}$. Two roads, one number, and the order of magnitude matches a real heater, which draws a few kilowatts.
Carry your answer forward, and check by a route avoiding it.
2§08.0 — combining two resistances, before any rule is stated●●○○○
This one is a trap, deliberately. Answer it with whatever instinct you have now, then look at the answer; the instinct that fails here is the single most expensive one in this whole section. A $2.00\ \Omega$ resistor and a $6.00\ \Omega$ resistor are joined so that both ends of each are attached to the same two terminals, and those terminals are then connected to a source.
Given
$R_1 = 2.00\ \Omega$ and $R_2 = 6.00\ \Omega$
both ends of each resistor are joined to the same two terminals
the pair is then treated as a single resistance
Find
(a) What single resistance behaves in the same way as the pair?
Hint 1/4
Ask a physical question rather than an algebraic one: with two paths open instead of one, is it easier or harder for charge to get across? The answer has to be smaller than one of the two numbers you were given.
Hint 2/4
When both ends of each resistor are on the same two terminals, both carry the same potential difference, the currents add, and the reciprocals of the resistances add.
Hint 3/4
With $R_1 = 2.00\ \Omega$ and $R_2 = 6.00\ \Omega$: $1/R = 1/2.00 + 1/6.00 = 4/6.00$.
Hint 4/4
Inverting that last line gives $R = 1.50\ \Omega$, smaller than either resistor, as it must be.
a common denominator keeps the two fractions exact, which matters because the last operation is an inversion and it magnifies any rounding done early
$$R = \frac{6.00}{4} = 1.50\ \Omega$$
the inversion is the step that is forgotten, so it is written on its own line rather than tacked onto the end of the previous one
Answer $$\boxed{\,R = 1.50\ \Omega\,}$$
Check
Independent check by a current argument at a chosen voltage, which never uses the formula: put $12.0\ \mathrm{V}$ across the pair. The first carries $6.00\ \mathrm{A}$, the second $2.00\ \mathrm{A}$, so the source delivers $8.00\ \mathrm{A}$ and sees $12.0/8.00 = 1.50\ \Omega$.
A parallel pair is smaller than its smallest branch; invert last.
3§08.0 — a charged capacitor, from the previous section●○○○○
The last of the three, and the only one that reaches back two sections. A $4.00\ \mathrm{\mu F}$ capacitor is charged until the potential difference across it is $50.0\ \mathrm{V}$.
of the three equivalent forms this one uses only the two given quantities, so an error in part a cannot propagate into part b
Answer $$\boxed{\,q = 200\ \mathrm{\mu C},\qquad U = 5.00\ \mathrm{mJ}\,}$$
Check
Independent check on the energy through the other form: $U = q^{2}/2C = (2.00\times10^{-4})^{2}/(8.00\times10^{-6}) = 5.00\times10^{-3}\ \mathrm{J}$, using the charge from part a rather than the voltage from the question.
Use the energy form built only from what the question gave.
Notation
symbol
reads as
means
watch out
$\varepsilon$
epsilon, the emf
the of a source, in volts: the energy per coulomb the source gives to charge passing through it
it is not a force despite the name, and it is not the voltage across the source's own terminals unless the current is zero
$r$
little r
the internal resistance of a source, in ohms; always in series with the emf
lower case r is reserved for internal resistance on this page; every other resistance is a capital R with a subscript
$V_{ab}$
V a b
the potential at point a minus the potential at point b, in volts
the order of the subscripts is the order of the subtraction, so $V_{ba} = -V_{ab}$
$I$
I
the current in a branch, in amperes, taken in the direction of the arrow drawn on that branch
a negative value is a legitimate answer and means the arrow was drawn the wrong way; it does not mean the arithmetic failed
$R_{\rm eq}$
R equivalent
the single resistance that would draw the same current from the same source as the whole network it replaces
it stands in for the network only as seen from outside; the individual currents inside still have to be recovered one at a time
$q(t)$
q of t
the charge on the positive plate of a capacitor at time t, in coulombs
lower case q is the running, time dependent value; capital $Q_0$ is a fixed starting value and $C\varepsilon$ is the final value it approaches
$\tau$
tau, the time constant
the product $RC$, in seconds, which sets the pace of every exponential on this page
$R$ here is the resistance the capacitor actually discharges through, which is not always the resistor that happens to be drawn next to it
$P$
P
power in watts: the rate a source delivers energy, or the rate a resistance turns it into heat
$P = I^2R$ and $P = V^2/R$ are the same statement only for a resistance; for a source the rate of energy conversion is $\varepsilon I$, and that is not $I^2r$
Conventions used here
Which way a current arrow points
Every current in these notes is a conventional current: the direction positive charge would move, which is out of the positive terminal of a source and through the external circuit. Inside a metal it is the electrons that actually move, in the opposite direction, and nothing on this page depends on that.
Assumed directions are allowed to be wrong
Before solving anything you draw an arrow on every branch and label it. The arrow is a guess. If the algebra returns a negative number, the magnitude is right and the real current runs the other way. You never go back and redraw: changing the arrow after writing the equations is how sign errors are manufactured.
Signs when travelling around a loop
Travelling through a resistance with the current arrow gives $\Delta V = -IR$, and against it gives $+IR$. Travelling through a source from its negative plate to its positive plate gives $+\varepsilon$, and the other way gives $-\varepsilon$, whatever the current is doing. Internal resistance is treated exactly like any other resistance, using the current in that branch.
What is ideal and what is not
Connecting wires have zero resistance and a switch has zero resistance when closed and infinite resistance when open. A battery is a source of emf with a resistance $r$ in series with it, and $r$ is zero only when the problem says so. An ideal ammeter has zero resistance and an ideal voltmeter infinite resistance; when a meter is given a resistance it is a circuit element like any other.
Steady state and time dependence
A direct-current circuit is one whose sources are constant. Everything settles to a steady state in which no charge accumulates anywhere, and the only element on this page that has a transient before it settles is the capacitor. In the steady state its branch carries no current at all.
Significant figures and constants
Answers are quoted to three significant figures, and intermediate values are carried at full precision and only rounded at the end. No physical constant beyond $e = 2.71828\ldots$ is needed anywhere on this page, and the elementary charge, when it appears, is $1.602\times10^{-19}\ \mathrm{C}$.
8.1Sources of emf, internal resistance, and the voltage you actually get
A source carries a resistance inside it, so the voltage at its terminals falls as soon as you draw current.
Everything so far has treated a battery as a fixed voltage that does not care what is attached to it. The dimming headlights say otherwise.
Solvable with what we have
Find a current from a voltage and a resistance.
Find a wire's resistance from its length, thickness and resistivity.
Find a power from a current and a voltage.
Not solvable yet
Say what current a real battery gives when a great deal is asked of it.
Explain why the lamps dim while the starter turns, then recover.
Explain why a battery that reads well doing nothing collapses under load.
A starter motor has a resistance of $0.0800\ \Omega$ and the battery is labelled $12.0\ \mathrm{V}$, so the current should be $I = V/R = 12.0/0.0800 = 150\ \mathrm{A}$ and the motor should receive $IV = 1800\ \mathrm{W}$. Measure it and you get about $120\ \mathrm{A}$, and the battery terminals read about $9.6\ \mathrm{V}$ while it is turning, not $12.0\ \mathrm{V}$.
Why it fails
The calculation assumes the terminals stay $12.0\ \mathrm{V}$ apart whatever is drawn from them, and that is the only assumption in it that can be wrong. Charge must be pushed through the chemistry inside the battery as well as through the motor outside, and that passage is not free.
DefinitionDefinition 8.1: emf, internal resistance and terminal voltage
Conditions
$\varepsilon$ is the energy per unit charge the source supplies to charge passing through it, in volts
$r$ is the internal resistance, in series with the emf and inseparable from it
$I$ is the current delivered by the source, taken as positive when it leaves the positive terminal
$$\boxed{\,V_{ab} = \varepsilon - Ir\,}$$
The voltage you can measure between the terminals is the full emf minus what is lost pushing the current through the source itself. Draw no current and you measure the emf exactly; draw a large current and you measure much less.
Proof
Model the source as an ideal seat of emf, which raises the potential of every coulomb passing through it by $\varepsilon$, in series with an ordinary resistance $r$.
Start at the negative terminal $b$ and walk through the source to the positive terminal $a$. The seat of emf raises the potential by $\varepsilon$.
The same walk passes through $r$ in the direction the current is flowing, which lowers the potential by $Ir$.
Adding the two changes gives the potential of $a$ above $b$: $V_{ab} = \varepsilon - Ir$.
Two limits check the result. With $I = 0$, nothing is lost inside and $V_{ab} = \varepsilon$: the terminal voltage of a battery sitting on the shelf is its emf. With the terminals joined by a perfect wire, $V_{ab} = 0$ and the current is $\varepsilon/r$, which is as much as the source can ever deliver.
The real source, opened up. Inside the dashed line are the two things you can never separate: the $\textcolor{#1f6feb}{\text{emf}}$, which is the full offer per coulomb, and the $\textcolor{#cf222e}{\text{internal resistance}}$, which takes its cut in proportion to the current. What reaches the $\textcolor{#1a7f37}{\text{external circuit}}$ is whatever is left between the terminals a and b.
Looks like this, but is not
A cell is labelled $1.50\ \mathrm{V}$, so a voltmeter across it must read $1.50\ \mathrm{V}$, and a $0.300\ \Omega$ lamp connected to it must carry $1.50/0.300 = 5.00\ \mathrm{A}$. Both come from one reasonable place: the label is the cell's voltage.
A voltmeter draws almost no current, so the $Ir$ term is negligible and its reading really is close to the emf. The lamp draws a great deal: with $r = 0.300\ \Omega$ the current is only $2.50\ \mathrm{A}$ and the lamp gets $0.750\ \mathrm{V}$. The label is a promise about what happens when you take nothing from the cell.
load R (Ω)
current I (A)
terminal voltage (V)
power in load (W)
power wasted in r (W)
100
0.0150
1.50
0.0224
0.0000671
3.00
0.455
1.36
0.620
0.0620
0.300
2.50
0.750
1.88
1.88
0.100
3.75
0.375
1.41
4.22
0
5.00
0
0
7.50
Read the third column downwards: the terminal voltage is not a property of the cell, it is a property of the cell together with whatever is attached to it, and it slides from the full emf all the way to zero. Read the last two columns across the middle row: when the load happens to equal the internal resistance, the load and the cell get exactly the same power, and that row is also where the power in the load is largest. Making the load smaller after that point makes the current bigger and the delivered power smaller, which is the opposite of what most people expect.
The starter motor, and where the missing volts went
A car battery has an emf of $12.0\ \mathrm{V}$ and an internal resistance of $0.0200\ \Omega$. It is connected to a starter motor whose resistance is $0.0800\ \Omega$. Find the current, the terminal voltage while the motor turns, the power delivered to the motor and the power wasted inside the battery.
Given
$\varepsilon = 12.0\ \mathrm{V}$
$r = 0.0200\ \Omega$
$R = 0.0800\ \Omega$ for the motor
Find
the current, the terminal voltage, and the two powers
with the current in hand the loss inside the battery is an ordinary Ohm's law drop, and it is subtracted rather than added because the current is leaving the positive terminal
Split the power into the useful part and the wasted part
the form $I^{2}R$ is chosen because the current is common to both parts, so the same number can be reused for the second line without recomputing anything
Independent check by energy bookkeeping, using none of the three lines above twice: the chemistry delivers $\varepsilon I = (12.0)(120) = 1440\ \mathrm{W}$, and the two powers found separately add to $1152 + 288 = 1440\ \mathrm{W}$. The order of magnitude is right too: a starter is about a kilowatt and a half, comparable to a kettle.
One addition, one division, one subtraction and two squares. The only place marks are lost is putting $R$ alone in the denominator.
Eighty per cent of the power reaches the motor and twenty per cent is wasted inside. That split is fixed by the ratio $R/(R+r)$ and by nothing else, so as a battery ages and its internal resistance climbs, the fraction that reaches the load falls even though the emf on the label has not moved.
Why the headlights dim while the engine is cranking
The same battery, $\varepsilon = 12.0\ \mathrm{V}$ and $r = 0.0200\ \Omega$, also feeds headlights of total resistance $4.00\ \Omega$. Find the power in the lamps with the lights on and the engine off, then find it again while the $0.0800\ \Omega$ starter motor is connected as well. Treat the filament as a fixed resistance.
Given
$\varepsilon = 12.0\ \mathrm{V}$ and $r = 0.0200\ \Omega$
lamps of total resistance $4.00\ \Omega$
starter of resistance $0.0800\ \Omega$, connected across the same terminals
with only the lamps attached the current is small, so the internal drop is only six hundredths of a volt and the terminal voltage is essentially the emf
the lamps and the starter are attached to the same two terminals, so they carry the same voltage and combine as a parallel pair; the tiny starter resistance dominates the result
the lamp resistance has not changed, so the whole effect comes through the terminal voltage, and the power depends on its square
Answer $$\boxed{\,P_{\rm lamps}: 35.6\ \mathrm{W}\ \to\ 22.9\ \mathrm{W}\quad(\text{a fall to } 64\%)\,}$$
Check
Independent consistency check on the branch currents, which the power calculation never used: the lamps take $9.56/4.00 = 2.39\ \mathrm{A}$ and the starter $9.56/0.0800 = 119.5\ \mathrm{A}$, and these add to $121.9\ \mathrm{A}$, the total current already found.
This is the hook, answered with numbers. Nothing happened to the lamps at all; the starter pulled the terminal voltage down by two and a half volts and the lamps merely reported it. Any two devices sharing one real source are coupled through $r$, and the coupling is invisible in a diagram that draws the battery as a perfect source.
Four cells in a torch: end to end, or side by side?
Four identical cells, each with $\varepsilon = 1.50\ \mathrm{V}$ and $r = 0.400\ \Omega$, are to drive a lamp of resistance $2.40\ \Omega$. Compare joining all four end to end, with the positive of each to the negative of the next, against joining all four side by side with all the positives together and all the negatives together.
Given
four cells, each $\varepsilon = 1.50\ \mathrm{V}$ and $r = 0.400\ \Omega$
lamp resistance $R = 2.40\ \Omega$
the two arrangements described
Find
the current in the lamp for each arrangement
SolutionEnd to end: the emfs add and so do the internal resistances
each coulomb passes through only one cell, so it collects one lot of energy; the four internal resistances are four identical parallel paths, so they combine to a quarter of one
the same single loop calculation, with the combined source
Answer $$\boxed{\,\text{end to end: } I = 1.50\ \mathrm{A};\qquad \text{side by side: } I = 0.600\ \mathrm{A}\,}$$
Check
Independent check on the side by side case by symmetry rather than by formula: the four identical cells must share the load current equally, so each supplies $0.150\ \mathrm{A}$ and loses $(0.150)(0.400) = 0.0600\ \mathrm{V}$ inside itself, leaving $1.44\ \mathrm{V}$ across the lamp; and $1.44/2.40 = 0.600\ \mathrm{A}$ as found.
End to end buys voltage, side by side buys the ability to deliver current without the terminal voltage collapsing. A torch does the first because it needs volts. Side by side is what you want when one cell already has enough emf and the internal resistance is what is limiting you, which is why a car has one heavy battery rather than eight small ones in a chain.
Checkpoint
§08.1 — the largest current a cell can ever give●●○○○
Thirty seconds. A cell has an emf of $1.50\ \mathrm{V}$ and an internal resistance of $0.300\ \Omega$. Someone joins its two terminals directly with a thick copper wire of negligible resistance.
Given
$\varepsilon = 1.50\ \mathrm{V}$
$r = 0.300\ \Omega$
the external resistance is zero
Find
(a) What current flows?
(b) What would a voltmeter across the terminals read while this is happening?
Hint 1/4
With the external resistance gone, ask what resistance is left in the loop. There is exactly one thing left, and it is inside the cell.
Hint 2/4
The loop current is $I = \varepsilon/(R+r)$, and the terminal voltage is $V_{ab} = \varepsilon - Ir$.
Hint 3/4
Here $\varepsilon = 1.50\ \mathrm{V}$, $r = 0.300\ \Omega$ and $R = 0$, so the denominator is $0 + 0.300$.
Hint 4/4
The current is $5.00\ \mathrm{A}$ and the terminal voltage is zero.
Independent check by energy: the chemistry delivers $\varepsilon I = 7.50\ \mathrm{W}$, and since nothing outside the cell has any resistance, all of it must be dissipated inside. Directly, $I^{2}r = (25.0)(0.300) = 7.50\ \mathrm{W}$, which agrees.
Internal resistance caps the current; the terminals read zero.
⚠ Putting only the external resistance in the denominator
the internal resistance is not drawn on most circuit diagrams, and what is not drawn is not added
wrong$$I = \frac{\varepsilon}{R}$$
right$$I = \frac{\varepsilon}{R+r}$$
⚠ Calling the emf the voltage across the load
the number on the battery is the most visible number in the problem, so it gets used for the most visible quantity
wrong$$V_{\rm load} = \varepsilon$$
right$$V_{\rm load} = V_{ab} = \varepsilon - Ir,\qquad \text{equal to } \varepsilon \text{ only when } I = 0$$
⚠ Reporting the power delivered by the source as the power delivered to the load
$\varepsilon I$ is the first product you can form from the given numbers, and it really is a power
wrong$$P_{\rm load} = \varepsilon I$$
right$$P_{\rm load} = I^{2}R = \varepsilon I - I^{2}r$$
8.2Resistors in series and in parallel: what is shared decides everything
Same current means add the resistances; same voltage means add the reciprocals and invert at the end.
The previous block put two resistances in one loop and added them without comment, and that quiet step is worth making explicit, because there is a second way to join two resistors and it obeys the opposite rule.
RuleResult 8.2: combining resistors
Conditions
Series: the same current passes through both, with no junction between them where current could leave
Parallel: both ends of each are joined to the same two conductors, so the same potential difference is across both
Every resistor is ohmic and the wires joining them have no resistance of their own
Along one road the resistances simply add, because the same current has to fight its way through each in turn and the voltages needed for each add up. Across two roads it is the reciprocals that add, because each road takes its own current at the same voltage and the currents add. A parallel combination is always smaller than the smallest resistor in it, and a series combination is always larger than the largest.
Proof
Series. The same current $I$ is in both, so the potential drops are $IR_1$ and $IR_2$. Walking through both drops the potential by $V = IR_1+IR_2 = I(R_1+R_2)$, and a single resistor drawing the same current from the same voltage would have $R = V/I = R_1+R_2$.
Parallel. The same potential difference $V$ is across both, so the currents are $V/R_1$ and $V/R_2$. Charge cannot pile up at the junction, so the current arriving is $I = V/R_1 + V/R_2$, and a single resistor drawing that current at that voltage would have $1/R = I/V = 1/R_1 + 1/R_2$.
Both derivations use only Ohm's law and one physical statement each: for series it is that the current has nowhere else to go, for parallel it is that both ends sit on the same two conductors. When you cannot say which of those two statements is true of a pair, the pair is neither, and neither formula applies.
A special case worth memorising: two in parallel give $R_{\rm par} = R_1R_2/(R_1+R_2)$, the product over the sum, which is quicker for a pair and wrong for three.
The two arrangements, labelled by the quantity they force to be shared. On the left one road, so the $\textcolor{#1f6feb}{\text{current}}$ is common and the voltages add; on the right two roads between the same pair of conductors, so the $\textcolor{#cf222e}{\text{potential difference}}$ is common and the currents add. Deciding which picture you are looking at is the whole job.
Looks like this, but is not
In the picture, $R_2$ and $R_3$ are drawn one above the other between the same two vertical wires, so they are in parallel and combine to $R_2R_3/(R_2+R_3)$. The drawing really does show them side by side, and the formula really is the one for a parallel pair.
Side by side on the page is not the test. The test is whether both ends of $R_2$ are joined to both ends of $R_3$ by conductors carrying nothing else. If a third wire leaves the node between them and carries current away, the two are not across the same pair of conductors and no combination rule applies to them at all. This is not a rare pathology: the standard bridge network has five resistors of which no two are in series and no two are in parallel, and it is exactly the situation the next block exists to handle. Before combining anything, trace each end of each resistor and check what else is attached there.
Two resistors in parallel, in series with a third
A $4.00\ \Omega$ and a $6.00\ \Omega$ resistor are connected in parallel with each other, and that combination is in series with a $2.00\ \Omega$ resistor across an ideal $12.0\ \mathrm{V}$ source. Find the equivalent resistance, the current from the source, and the current in each of the three resistors.
Given
$R_1 = 4.00\ \Omega$ and $R_2 = 6.00\ \Omega$ in parallel
the innermost group is collapsed first because the outer series addition cannot be done until the pair has become a single number; the product over sum shortcut is used because there are exactly two of them
now there is one road through two resistances in turn, so they add
Total current, from the source's point of view
$$I = \frac{12.0}{4.40} = 2.73\ \mathrm{A}$$
the source is ideal, so the whole $12.0\ \mathrm{V}$ is across the network and no internal resistance appears in the denominator
Work back outwards, carrying what each grouping shares
$$V_3 = IR_3 = (2.727)(2.00) = 5.45\ \mathrm{V}$$
the series resistor carries the whole current, so it is the one element whose voltage can be written down immediately
$$V_{12} = 12.0 - 5.45 = 6.55\ \mathrm{V}$$
the two voltages must add to the source voltage, so subtracting is safer than recomputing $IR_{12}$, which would hide an arithmetic error in the previous line
Independent check with a sum that was never used in the solution: the two branch currents must add up to the current that arrived, and $1.64+1.09 = 2.73\ \mathrm{A}$. Note also that the smaller resistor took the larger current, in the ratio $6:4$, which is the reverse of the resistance ratio as it must be.
Two combinations inwards, three steps outwards. The outward journey is where most of the marks are and where most people stop.
The pattern to carry away is that the inward journey gives you one number and tells you nothing about the inside, and the outward journey is where the question is actually answered. Every network question on this page has the same two halves.
Two hundred watt bulbs in series give fifty watts, not two hundred
Two lamps, each rated $100\ \mathrm{W}$ at $230\ \mathrm{V}$, are connected first in parallel across a $230\ \mathrm{V}$ supply and then in series across the same supply. Find the power in each lamp in each case, treating the filament resistance as constant.
Given
each lamp is rated $100\ \mathrm{W}$ at $230\ \mathrm{V}$
a rating is a pair of numbers that only holds at the rated voltage; the resistance is the thing that stays with the lamp when the voltage changes, so we convert to it first
Independent check on the series case through the voltage instead of the current: the two identical lamps split the supply equally, so each has $115\ \mathrm{V}$, and $V^{2}/R = 13225/529 = 25.0\ \mathrm{W}$. Both routes agree, and the factor of four is the square of the factor of two in the voltage.
Halving the voltage across a lamp quarters its power, because power goes as the square. That is why old Christmas lights wired end to end are so dim, and why one lamp of a series pair failing takes the other with it: the current stops everywhere at once.
Checkpoint
§08.2 — what a third parallel branch does●●○○○
Thirty seconds, and the second part is the one that catches people. Two $4.00\ \Omega$ resistors are connected in parallel across an ideal $12.0\ \mathrm{V}$ source. A third $4.00\ \Omega$ resistor is then connected in parallel with them.
Given
two $4.00\ \Omega$ resistors in parallel across an ideal $12.0\ \mathrm{V}$ source
a third $4.00\ \Omega$ resistor added in parallel
the source is ideal, so its terminal voltage stays at $12.0\ \mathrm{V}$
Find
(a) What is the equivalent resistance before and after?
(b) Does the current in the first resistor change when the third is added?
Hint 1/4
For part b, ask what the current in the first resistor is actually determined by. Write down the two quantities it depends on and then ask whether either of them was touched.
Hint 2/4
For $n$ identical resistors in parallel, $R_{\rm par} = R/n$; and the current in any one branch is the voltage across that branch divided by its own resistance.
Hint 3/4
Here $R = 4.00\ \Omega$ with $n$ going from two to three, and the source holds $12.0\ \mathrm{V}$ across the branches throughout.
Hint 4/4
The equivalent resistance falls from $2.00\ \Omega$ to $1.33\ \Omega$, and the current in the first resistor does not change at all.
Independent check by adding the branch currents rather than using the equivalent resistance: three identical branches at $3.00\ \mathrm{A}$ give $9.00\ \mathrm{A}$, and $12.0/9.00 = 1.33\ \Omega$, the equivalent resistance found the other way.
A branch current knows only its own voltage and its own resistance.
⚠ Adding parallel resistors as though they were in series
addition is the rule that gets remembered, and the two pictures are drawn a few centimetres apart in every book
⚠ Splitting the current equally between unequal parallel branches
the branches look symmetric in the drawing and the word parallel suggests sameness
wrong$$I_1 = I_2 = \tfrac12 I$$
right$$I_i = \frac{V}{R_i},\qquad \text{so } I_1R_1 = I_2R_2,\ \text{equal only if } R_1 = R_2$$
8.3Reduce, then expand: the whole circuit from one equivalent resistance
Collapse the network inwards to one number, get the total current, then walk back outwards carrying whatever each grouping shares.
We now have the two combination rules and a source that is honest about its internal resistance, and putting them together turns a picture full of unknowns into a procedure with a fixed number of steps.
TheoremResult 8.3: the power balance of a complete circuit
Conditions
Every source in the circuit is included, each with its own current $I_k$ through it
Every resistance is included, internal resistances among them, each with the current $I_j$ actually in it
The circuit is in a steady state, so nothing is storing energy
The rate at which all the sources hand out energy equals the rate at which all the resistances turn it into heat. Nothing else happens in a steady direct-current circuit, so if the two sides of this equation disagree, one of the currents is wrong.
Proof
Take the loop equation for the single loop case, $\varepsilon = I(R+r)$, and multiply both sides by the current $I$.
The left side becomes $\varepsilon I$, the energy per coulomb given out times the coulombs per second passing, which is the rate the source converts chemical energy into electrical energy.
The right side becomes $I^{2}R + I^{2}r$, the rate the two resistances heat up.
For a network the same argument runs branch by branch: multiply each loop equation by its own current and add. Every internal node contributes terms that cancel in pairs, because the junction rule says the currents arriving equal the currents leaving.
The result is worth more as a check than as a tool. It uses all the currents at once, and it uses them in a combination that none of the equations you solved involved, so an error in any single branch shows up as a mismatch.
One lap around the circuit of the first worked example, plotted as potential against position. The $\textcolor{#1f6feb}{\text{source}}$ lifts every coulomb by $12.0\ \mathrm{V}$ and each $\textcolor{#cf222e}{\text{resistance}}$ takes part of it back, in proportion to its own resistance and the current in it. You end the lap exactly where you started, which is the loop rule drawn rather than stated.
Looks like this, but is not
The parallel block came to $4.00\ \Omega$ and the total current is $1.20\ \mathrm{A}$, so the power in the $12.0\ \Omega$ resistor inside that block is $I^{2}R = (1.20)^{2}(12.0) = 17.3\ \mathrm{W}$. Every symbol has been given a value from the problem and the formula is the correct one for a resistor.
The whole circuit is only receiving $\varepsilon I = 14.4\ \mathrm{W}$, so a single resistor inside it cannot be dissipating $17.3\ \mathrm{W}$; the answer is impossible before it is even checked. The error is that $I$ in $I^{2}R$ must be the current in that resistor, and the $12.0\ \Omega$ branch carries only a third of the total. This is the reason the outward journey exists: an equivalent resistance is a statement about the network seen from outside, and the moment you ask about one element inside it you need that element's own current or its own voltage.
A real battery, one series resistor and a parallel pair, solved completely
A battery of emf $12.0\ \mathrm{V}$ and internal resistance $0.500\ \Omega$ drives a $5.50\ \Omega$ resistor in series with a parallel pair made of $12.0\ \Omega$ and $6.00\ \Omega$. Find the current from the battery, the terminal voltage, the current and voltage for every resistor, and the power dissipated in each.
the internal resistance is included here because the current from the battery passes through it exactly as it passes through the others; leaving it out is the most common error in this step
subtracting from the terminal voltage uses the loop, which is an extra constraint; computing $IR_{23}$ instead would give the same number without testing anything
Independent check by the power balance, which uses every number found and none of the equations solved: the source delivers $\varepsilon I = (12.0)(1.20) = 14.4\ \mathrm{W}$, and the four dissipations add to $0.720+7.92+1.92+3.84 = 14.4\ \mathrm{W}$. A second, cheaper check: the branch currents add, $0.400+0.800 = 1.20\ \mathrm{A}$.
Two combinations in, one division, five lines out and four powers. Roughly a third of the work is the inward journey and two thirds is the outward one.
Notice which quantity was carried at each stage of the outward journey: the current along the single road, then the voltage across the parallel block. That alternation is the whole method, and it is the same in every network no matter how many layers deep it goes.
Switch off one branch and the other one gets brighter
In the same circuit, $\varepsilon = 12.0\ \mathrm{V}$, $r = 0.500\ \Omega$, $R_1 = 5.50\ \Omega$ in series with $R_2 = 12.0\ \Omega$ parallel to $R_3 = 6.00\ \Omega$, a switch now disconnects $R_3$ completely. Find the new current from the battery and the new power in $R_2$, and compare both with the values before the switch was opened.
Given
the circuit of the previous example
$R_3 = 6.00\ \Omega$ is disconnected, leaving its branch open
before the change: $I = 1.20\ \mathrm{A}$, $I_2 = 0.400\ \mathrm{A}$, $P_2 = 1.92\ \mathrm{W}$
Find
the new total current and the new power in the surviving branch
Independent check on the new state with the loop rather than with the powers: $0.333 + 3.67 + 8.00 = 12.0\ \mathrm{V}$, the three drops across $r$, $R_1$ and $R_2$ adding to the emf as they must.
The total current fell and one particular resistor got hotter, and there is no contradiction: the missing branch stopped stealing voltage from the survivor. Whenever a question asks what happens to one element after a change, resist answering from the total; find that element's own voltage or its own current in the new circuit.
Checkpoint
§08.3 — a bulb in series with a parallel pair●●○○○
Thirty seconds, and it is a prediction rather than a full calculation. An ideal $12.0\ \mathrm{V}$ source drives a $2.00\ \Omega$ resistor in series with two $8.00\ \Omega$ resistors that are in parallel with each other.
Given
ideal source, $12.0\ \mathrm{V}$
$R_1 = 2.00\ \Omega$ in series
two $8.00\ \Omega$ resistors in parallel with each other
Find
(a) Find the total current.
(b) Find the current in one of the $8.00\ \Omega$ resistors.
Hint 1/4
Two journeys: collapse the pair and add, to get the total; then hand the pair the voltage it shares and split from there.
Hint 2/4
Identical resistors in parallel give $R/n$, series resistances add, and each branch of a parallel pair obeys $I_i = V/R_i$ on its own.
Hint 3/4
With $R_1 = 2.00\ \Omega$ and two $8.00\ \Omega$ branches: $R_{\rm par} = 4.00\ \Omega$, so $R_{\rm eq} = 6.00\ \Omega$ across $12.0\ \mathrm{V}$.
Hint 4/4
The total current is $2.00\ \mathrm{A}$ and each parallel branch takes half of it, $1.00\ \mathrm{A}$.
Independent check by the power balance: the source gives $(12.0)(2.00) = 24.0\ \mathrm{W}$, while $I^{2}R_1 = 8.00\ \mathrm{W}$ and the two branches give $(1.00)^{2}(8.00) = 8.00\ \mathrm{W}$ each, and $8.00+8.00+8.00 = 24.0\ \mathrm{W}$.
Inwards to one resistance, outwards carrying current then voltage.
⚠ Using the total current for a resistor inside a parallel branch
the total current is the number most recently written down, and $I^{2}R$ does not say which $I$ it wants
⚠ Treating an open branch as a very large resistance in a series sum
an open switch is described as infinite resistance, and infinity is then put into the formula literally
wrong$$R_{23} = \frac{(12.0)(\infty)}{12.0+\infty}\ \text{written as a large number}$$
right$$\text{an open branch carries no current: delete it, so } R_{23} \to 12.0\ \Omega$$
8.4Kirchhoff's two rules, and the four signs you have to get right
Charge does not pile up at a junction, and a round trip returns you to the same potential.
The counterexample two blocks ago promised a network in which no two resistors are in series and no two are in parallel, and for that network the whole reduce and expand method has nothing to grip on.
TheoremTheorem 8.4: Kirchhoff's junction and loop rules
Conditions
The circuit is in a steady state, so no charge is accumulating anywhere
Currents are labelled with assumed directions, which are allowed to be wrong
Every element in a chosen loop is included, internal resistances among them
At any junction, everything that arrives leaves, because charge is conserved and nothing can build up at a point in a wire. Around any closed path, the potential changes add to zero, because getting back to where you started means getting back to the potential you started at.
Proof
The junction rule is conservation of charge and nothing else. If more charge arrived at a point than left it, charge would accumulate there, the accumulated charge would create a field opposing further arrivals, and the circuit would not be in a steady state. In the steady state the two sums are equal.
The loop rule is the statement that electric potential is a single valued function of position. Walking from a point back to the same point, the total change in potential must be zero, because the potential at that point is one number.
This is why the loop rule works for any loop you care to draw, including loops with no source in them and loops that cut through the middle of a network.
The two rules are not new physics. Everything on this page before them is a special case: for two resistors in series the junction rule says the current is the same in both, and for two in parallel the loop rule around the pair says their voltages are equal.
The four cases, which is all there is to remember. For a $\textcolor{#cf222e}{\text{resistance}}$ the sign depends on whether you are walking with or against the current arrow; for a $\textcolor{#1f6feb}{\text{source}}$ it depends only on which plate you meet first, and the current is irrelevant. Walking direction is shown by the dashed arrow underneath each element.
Looks like this, but is not
This network has three loops in it, so I can write three loop equations, and with the junction equation that gives four equations for three unknown currents. More equations can only help. Drawing the third loop around the outside is easy and the equation it gives is perfectly true.
The outer loop equation is the sum of the two inner ones, so it carries no information that they do not already carry. Substituting the other two into it produces $0 = 0$, and a student who does not notice will chase an algebra error that does not exist. The count is fixed: a network with $n$ unknown currents needs exactly $n$ independent equations, and you get them by taking one fewer junction equation than there are junctions, and then just enough loops so that each loop you choose contains at least one branch that no previous loop contained.
One loop, two batteries facing each other, and one of them is being charged
A battery of emf $12.0\ \mathrm{V}$ and internal resistance $0.300\ \Omega$ is connected to a second battery of emf $6.00\ \mathrm{V}$ and internal resistance $0.200\ \Omega$ with their positive terminals facing each other through a $5.50\ \Omega$ resistor. Find the current, the terminal voltage of each battery, and say what is happening to the energy.
walking round in the direction the larger emf pushes, we meet battery one from minus to plus and battery two from plus to minus, which is why the second emf enters with the opposite sign
only the difference of the emfs drives the loop, because the smaller one is pushing the other way, while all three resistances still oppose the flow and so all three are added
Terminal voltage of each, which now differ in sign
current is being pushed into the positive terminal of battery two, so the internal drop is on the way in and the terminal voltage is above its emf; this is the only situation in which that happens
the product of emf and current is a rate of conversion in both cases; the direction of the current relative to the terminals decides which way the conversion runs
Independent check by the power balance: $12.0\ \mathrm{W}$ leaves the first battery, $6.00\ \mathrm{W}$ is stored chemically in the second and $6.00\ \mathrm{W}$ becomes heat, and $6.00+6.00 = 12.0\ \mathrm{W}$. Half of what the big battery gives up ends up in the small one, and half is wasted.
This is what a battery charger is. Note the sign of the correction: a terminal voltage above the emf is the signature of a source being charged, and it is not a mistake in your algebra when it happens.
A two loop network with a source in each of two branches
Two nodes, $a$ at the top and $b$ at the bottom, are joined by three branches. The first contains a $12.0\ \mathrm{V}$ source in series with $2.00\ \Omega$, the second a $6.00\ \mathrm{V}$ source in series with $3.00\ \Omega$, and the third contains only a $6.00\ \Omega$ resistor. Both sources have their positive terminals towards $a$. Find the current in each branch.
Given
branch 1: $\varepsilon_1 = 12.0\ \mathrm{V}$ with $R_1 = 2.00\ \Omega$, positive terminal towards $a$
branch 2: $\varepsilon_2 = 6.00\ \mathrm{V}$ with $R_2 = 3.00\ \Omega$, positive terminal towards $a$
branch 3: $R_3 = 6.00\ \Omega$ alone
Find
the three branch currents, with their true directions
SolutionLabel first, and count what you need
$$I_1,\ I_2\ \text{assumed from } b \text{ to } a;\quad I_3 \text{ assumed from } a \text{ to } b$$
three unknown currents means three independent equations; with two junctions we take one junction equation and two loop equations
$$I_1+I_2 = I_3$$
the junction rule at node a, with two arrows arriving and one leaving; the equation at node b is the same statement and would add nothing
Two loops, each with a branch the other one lacks
$$12.0-2.00I_1-6.00I_3 = 0$$
the left loop, walking up through the first source from minus to plus and then down through $R_3$ along $I_3$, so the emf enters positive and both resistor terms negative
$$6.00-3.00I_2-6.00I_3 = 0$$
the right loop, walked the same way; choosing loops that each contain the middle branch is what couples the two equations together
Solve, and let the sign speak
$$I_1 = 6.00-3.00I_3,\qquad I_2 = 2.00-2.00I_3$$
each loop equation is solved for its own branch current so that the junction equation becomes a single equation in $I_3$
the negative sign is the answer, not an error: the current in branch two really runs from $a$ to $b$, so the six volt source is being charged by the twelve volt one
Independent check by computing the potential difference $V_{ab}$ along all three branches separately, which the solution never did: through branch three, $V_{ab} = I_3R_3 = 8.00\ \mathrm{V}$; through branch one, $12.0-(2.00)(2.00) = 8.00\ \mathrm{V}$; through branch two, $6.00-(3.00)(-0.667) = 8.00\ \mathrm{V}$. Three routes between the same two points, one number.
Three equations, one substitution and one division. Every sign in them was fixed before any number was written down.
The general shape of these problems: label, count, write exactly as many independent equations as unknowns, solve, and then read the signs. If you find yourself needing a fourth equation, you have used two loops that share all their branches.
Checkpoint
§08.4 — counting the equations before writing any●●○○○
Thirty seconds, and it is about bookkeeping rather than about physics. A network has four branches meeting at two junctions, and three independent closed loops can be traced in the drawing.
Given
four branches, so four unknown branch currents
two junctions
three closed loops can be drawn
Find
(a) How many independent junction equations are there?
(b) How many loop equations must you then write?
Hint 1/4
Count the unknowns first. Every independent equation you write must be paid for by an unknown, and the two rules between them must supply exactly that many.
Hint 2/4
With $j$ junctions, exactly $j-1$ junction equations are independent; the last one is the sum of the others. The rest of the equations must come from loops, each chosen to contain a branch that no earlier loop contained.
Hint 3/4
Here there are four unknown currents and two junctions, so $j-1 = 1$.
Hint 4/4
One junction equation and three loop equations, four in total for four unknowns.
Show solutionCount unknowns and independent junction equations
the last junction equation is the sum of all the others, so it is never new information no matter how many junctions there are
Fill the shortfall with loops
$$N_{\rm loop} = 4-1 = 3$$
the loops are chosen, not counted from the picture: each new loop must contain a branch that no earlier loop contained, otherwise its equation is a combination of theirs
Independent check against the simplest case: a single loop with one source has one branch current, one junction count of zero, and needs one loop equation. The rule gives $j-1 = 0$ junction equations and $1-0 = 1$ loop equation, which is what everybody does by instinct.
Count the unknown currents first; junctions give $j-1$ and loops the rest.
⚠ Writing minus IR whichever way you are walking
the drop across a resistor is remembered as a fact about resistors, when it is a fact about direction of travel
wrong$$\text{walking against } I:\ \Delta V = -IR$$
right$$\text{walking against } I:\ \Delta V = +IR$$
⚠ Adding the outer loop as an extra equation
it is a genuine loop and its equation is genuinely true, so it looks like free information
A resistance limits how fast charge arrives, so the capacitor fills exponentially at a pace set by RC.
Every circuit so far settled the instant the switch closed, and adding one capacitor changes that: now there is a quantity that cannot jump, so the circuit has a history.
TheoremResult 8.5: charging a capacitor through a resistance
Conditions
One loop containing a constant source $\varepsilon$, a total resistance $R$ and a capacitance $C$
The capacitor is uncharged at $t = 0$, when the switch is closed
$R$ is the total resistance in the loop, including any internal resistance
The charge climbs towards the value the source would eventually force onto the capacitor, and the gap remaining shrinks by the same factor in every equal interval of time. The current does the opposite: it is largest at the very first instant, when the capacitor is empty and cannot push back, and it dies away as the capacitor fills.
Proof
Walk round the loop at some instant when the charge is $q$ and the current is $I$. The source lifts by $\varepsilon$, the resistance drops $IR$, and the capacitor drops $q/C$, so $\varepsilon - IR - q/C = 0$.
The current is the rate at which charge is arriving on the plate, $I = dq/dt$. Substituting turns the loop rule into a differential equation: $R\,dq/dt = \varepsilon - q/C$.
Write the right side over a common denominator, $R\,dq/dt = (C\varepsilon - q)/C$, and separate the variables: $\dfrac{dq}{C\varepsilon-q} = \dfrac{dt}{RC}$.
Integrate both sides from the closing of the switch: the left gives $-\ln(C\varepsilon-q)$ evaluated between $0$ and $q$, the right gives $t/RC$.
Rearranging, $\ln\dfrac{C\varepsilon-q}{C\varepsilon} = -\dfrac{t}{RC}$, and taking exponentials gives the charging law.
Differentiate it to get the current, $I = dq/dt = (\varepsilon/R)e^{-t/RC}$, and check the two ends. At $t=0$ the capacitor is empty, so it opposes nothing and the current is $\varepsilon/R$, exactly as if the capacitor were a plain wire. After a long time the current has stopped and $q \to C\varepsilon$, exactly as if the capacitor were a break in the circuit.
The charge on the plate as a fraction of its final value, against time in units of the time constant. The $\textcolor{#1f6feb}{\text{initial slope}}$ is the picture of the time constant: if the capacitor kept filling at the rate it starts with, it would be full after exactly one $\tau$. It does not, because the current falls as the charge grows, so it arrives at $63\%$ instead.
Looks like this, but is not
The circuit has a $20.0\ \mathrm{k\Omega}$ resistor and a $5.00\ \mathrm{\mu F}$ capacitor, so the capacitor is fully charged after $\tau = 0.100\ \mathrm{s}$. The time constant really is the natural time scale of the circuit, and it really does have units of seconds.
After one time constant the capacitor holds $63\%$ of its final charge, not all of it. The exponential never reaches its final value at any finite time; each further time constant removes another factor of $e$ from what is still missing, leaving $37\%$ missing after one, $14\%$ after two and $0.7\%$ after five. The working rule is that five time constants is fully charged for any practical purpose, and the honest statement is that the time constant is the time to close $63\%$ of the remaining gap, whenever you start counting.
t (s)
t / τ
q (μC)
I (mA)
charge still missing (μC)
0
0
0
0.600
60.0
0.100
1
37.9
0.221
22.1
0.200
2
51.9
0.0812
8.1
0.300
3
57.0
0.0299
3.0
0.500
5
59.6
0.00404
0.4
Read the third and fourth columns together: whatever fraction of the charge is still missing is the same fraction of the initial current that is still flowing. That is not a coincidence, it is the loop rule, since the missing charge and the current are the two things that must add up to the emf. Read the third column downwards and notice that the gaps shrink by the same factor each row, which is what an exponential means: from 22.1 to 8.1 to 3.0 microcoulombs missing, each about $0.37$ of the one before.
Twelve volts filling five microfarads through twenty kilohms
A $12.0\ \mathrm{V}$ source of negligible internal resistance, a $20.0\ \mathrm{k\Omega}$ resistor and an uncharged $5.00\ \mathrm{\mu F}$ capacitor are joined in one loop, and the switch is closed at $t = 0$. Find the time constant, the final charge, the current at the instant of closing, the charge and current at $t = 0.250\ \mathrm{s}$, and the time at which the capacitor is $90\%$ charged.
Independent check that the loop rule holds at $t = 0.250\ \mathrm{s}$, using both answers at once: the capacitor voltage is $q/C = 55.1/5.00 = 11.0\ \mathrm{V}$ and the resistor voltage is $IR = (4.93\times10^{-5})(2.00\times10^{4}) = 0.99\ \mathrm{V}$, and the two add to $12.0\ \mathrm{V}$, the emf. If either answer were wrong the sum would miss.
Three constants, one exponential, one logarithm. Every quantity after the first line is a pure multiple of one of those three constants.
Carry away the habit of computing $\tau$, $Q_f$ and $I_0$ before touching the time in the question. Every later number is one of those three multiplied by an exponential factor, and the fractions $0.63$, $0.86$, $0.95$ and $0.993$ at one, two, three and five time constants are worth knowing by heart.
A capacitor in a network, long after the switch was closed
An ideal $24.0\ \mathrm{V}$ source drives a $3.00\ \mathrm{k\Omega}$ resistor in series with a $6.00\ \mathrm{k\Omega}$ resistor. A branch containing a $2.00\ \mathrm{k\Omega}$ resistor in series with a $2.00\ \mathrm{\mu F}$ capacitor is connected across the $6.00\ \mathrm{k\Omega}$ resistor. Long after the switch was closed, find the current in every resistor and the charge on the capacitor.
Given
$\varepsilon = 24.0\ \mathrm{V}$, ideal
$R_1 = 3.00\ \mathrm{k\Omega}$ in series with $R_2 = 6.00\ \mathrm{k\Omega}$
a branch of $R_3 = 2.00\ \mathrm{k\Omega}$ and $C = 2.00\ \mathrm{\mu F}$ across $R_2$
the circuit has been running for a long time
Find
the steady currents and the charge on the capacitor
SolutionDecide what the capacitor is doing, before any arithmetic
$$I_C = \frac{dq}{dt} = 0 \ \text{in the steady state}$$
charge has stopped arriving on the plates, otherwise the state would still be changing, so the capacitor branch behaves as a break in the circuit
$$V_{R_3} = I_CR_3 = 0$$
no current in that branch means no drop across the resistor in it, which is the step that makes $R_3$ irrelevant to the final answer
going along the branch from one end to the other, the only two elements are $R_3$, which drops nothing, and the capacitor, so the whole $16.0\ \mathrm{V}$ appears across the capacitor
Independent check on the loop that contains the capacitor, which the solution used only in part: going from the source through $R_1$, then along the capacitor branch, gives $24.0-(2.667)(3.00)-0-16.0 = 0\ \mathrm{V}$, so that loop closes. Order of magnitude: tens of microcoulombs on a couple of microfarads at tens of volts is right.
Two rules to carry away, both about the ends of the story rather than the middle. Long after the switch closes, a capacitor branch is a break and carries no current. At the very instant the switch closes on an uncharged capacitor, it is the opposite: the capacitor has no voltage across it and behaves like a plain wire. Most exam questions about capacitors in networks ask about one of those two instants.
Checkpoint
§08.5 — reading the pace of a charging circuit●●○○○
Thirty seconds and two lines. A $9.00\ \mathrm{V}$ source of negligible internal resistance charges an uncharged $20.0\ \mathrm{\mu F}$ capacitor through a $100\ \mathrm{k\Omega}$ resistor.
(b) Find the current the instant the switch is closed.
Hint 1/4
Neither part needs the exponential. One is a product of the two circuit elements, the other is the circuit at the single instant when the capacitor is not yet pushing back.
Hint 2/4
$\tau = RC$, and at $t = 0$ an uncharged capacitor has no voltage across it, so the current is $\varepsilon/R$.
Hint 3/4
Here $R = 1.00\times10^{5}\ \Omega$, $C = 2.00\times10^{-5}\ \mathrm{F}$ and $\varepsilon = 9.00\ \mathrm{V}$.
Hint 4/4
The time constant is $2.00\ \mathrm{s}$ and the initial current is $90.0\ \mathrm{\mu A}$.
both quantities are put into base units first, because a kilohm times a microfarad is a millisecond and mixing the prefixes is the standard way to lose three orders of magnitude here
Independent consistency check on the units and the scale: the final charge is $C\varepsilon = 180\ \mathrm{\mu C}$, and dividing it by the initial current gives $180/90.0 = 2.00\ \mathrm{s}$, the time constant again. That ratio is $C\varepsilon/(\varepsilon/R) = RC$ for any such circuit, so the two answers had to be consistent in this way.
Work in ohms and farads; an uncharged capacitor starts as wire.
⚠ Multiplying the emf by the exponential bracket instead of the final charge
the emf is the number in the question and the final charge is one you have to compute first
wrong$$q = \varepsilon\left(1-e^{-t/RC}\right)$$
right$$q = C\varepsilon\left(1-e^{-t/RC}\right)$$
⚠ Giving the current the same bracket as the charge
the two laws are written one under the other and the bracket is the visually memorable part
With no source in the loop the capacitor drives its own current and falls by the same factor in every equal interval.
Take the same loop, remove the source, and the capacitor is now the only thing in the circuit with any energy in it, which turns the same differential equation into a pure decay.
TheoremResult 8.6: discharging, and the energy that never arrives
Conditions
A capacitor holding $Q_0$ is connected at $t=0$ to a resistance $R$, with no source in the loop
$R$ is the total resistance in the discharge path, which need not be the resistor used to charge it
For the energy statement: the capacitor is charged from a constant source through a resistance, starting empty
During a discharge the charge falls by a fixed factor in every equal stretch of time, and the pace is the same product of resistance and capacitance as before. And whenever a capacitor is charged from a constant source, exactly half the energy the source hands out is stored and exactly half is dissipated in the resistance, no matter what the resistance is.
Proof
With no source, the loop rule says the capacitor's voltage drives the current through the resistance: $q/C = IR$.
The charge on the plate is now falling, so the current out of the capacitor is $I = -dq/dt$. Substituting gives $q/C = -R\,dq/dt$.
Separate and integrate: $dq/q = -dt/RC$, so $\ln(q/Q_0) = -t/RC$ and $q = Q_0e^{-t/RC}$.
For the energy statement, integrate the heating over the whole charge: the current is $(\varepsilon/R)e^{-t/RC}$, so the rate of heating is $(\varepsilon^{2}/R)e^{-2t/RC}$.
Integrating that from zero to infinity multiplies it by $RC/2$, giving $\tfrac12 C\varepsilon^{2}$ of heat.
Compare with the two other quantities. The source delivered $Q_f\varepsilon = C\varepsilon^{2}$, and the capacitor kept $\tfrac12 C\varepsilon^{2}$. The resistance is nowhere in the answer, which is the surprising part: making the resistance smaller charges the capacitor faster but wastes exactly the same energy.
The discharge, with two ways of reading the same curve marked on it. After one $\textcolor{#cf222e}{\text{time constant}}$ what is left is $0.37$ of what you started with, and the $\textcolor{#1f6feb}{\text{half}}$ is reached earlier, at $0.69\,\tau$. Equal steps along the time axis always multiply the remaining charge by the same factor, which is the one property that defines an exponential.
Looks like this, but is not
The capacitor was charged through a $20.0\ \mathrm{k\Omega}$ resistor, so it discharges with $\tau = 0.100\ \mathrm{s}$. The time constant of the charging circuit really was $0.100\ \mathrm{s}$, and the capacitance has certainly not changed.
The time constant belongs to the loop the current is actually flowing round, not to the capacitor. If the capacitor is disconnected from the charging resistor and dumped through a $50.0\ \Omega$ lamp, the discharge constant is $(50.0)(5.00\times10^{-6}) = 2.50\times10^{-4}\ \mathrm{s}$, four hundred times faster, which is exactly how a camera flash works: charge slowly through a large resistance, discharge quickly through a small one. Before writing $\tau = RC$, ask which $R$ the charge is about to move through.
Dumping sixty microcoulombs through twenty kilohms
The capacitor of the previous block, $C = 5.00\ \mathrm{\mu F}$ holding $Q_0 = 60.0\ \mathrm{\mu C}$, is disconnected from the source and connected across the same $20.0\ \mathrm{k\Omega}$ resistor at $t = 0$. Find the initial current, the charge and current at $t = 0.150\ \mathrm{s}$, the time to fall to a tenth of the starting charge, and the total heat produced.
in a discharge both the charge and the current carry the same bare exponential, with no bracket anywhere, because both are decaying from their starting values
falling to a tenth takes 2.30 time constants, the same number as rising to nine tenths did, because both are the same statement about the missing fraction
Independent check on the heat by integrating the current instead of using the stored energy: the heating rate starts at $I_0^{2}R = 7.20\ \mathrm{mW}$ and decays with time constant $\tau/2 = 0.0500\ \mathrm{s}$, so the total is $(7.20\ \mathrm{mW})(0.0500\ \mathrm{s}) = 360\ \mathrm{\mu J}$, which matches.
The discharge law has no bracket in it and the charging law has one in the charge but not in the current. If you can remember which quantities start large and which start at zero, you can rebuild all four expressions from the two exponentials without memorising them separately.
Charging always wastes exactly half, whatever the resistor
An uncharged $5.00\ \mathrm{\mu F}$ capacitor is charged fully from a $12.0\ \mathrm{V}$ source through a resistor. Find the total energy the source delivers, the energy the capacitor ends up holding, and the energy dissipated in the resistor. Then repeat the argument for a resistor ten times smaller and say what changes.
Given
$C = 5.00\ \mathrm{\mu F}$, initially uncharged
$\varepsilon = 12.0\ \mathrm{V}$, constant
charged through a resistance $R$, then through $R/10$
every coulomb that leaves the source crosses the full emf, whatever the capacitor is doing at the time, so the total is simply the final charge times the emf and not half of it
Independent check by the integral rather than by subtraction: the heating rate is $(\varepsilon^{2}/R)e^{-2t/\tau}$, whose time integral is $(\varepsilon^{2}/R)(\tau/2) = \tfrac12 C\varepsilon^{2} = 360\ \mathrm{\mu J}$. The $R$ cancels between the height of the curve and its width, which is the reason the answer cannot depend on it.
This is a genuinely surprising result and it is worth stating plainly: you cannot charge a capacitor from a constant source through any resistance and keep more than half the energy. Charging it in stages, or through something that is not a plain resistance, is how that limit is beaten, and neither is on this page.
Checkpoint
§08.6 — half life and time constant are not the same number●●○○○
Thirty seconds. A capacitor discharging through a resistor is observed to lose half of its charge in $3.00\ \mathrm{s}$.
Given
the charge falls to one half in $3.00\ \mathrm{s}$
the discharge is through a fixed resistance
the capacitance is constant
Find
(a) What is the time constant?
(b) How long until only a quarter of the original charge is left?
Hint 1/4
Part b needs no exponential at all if you notice what a quarter is in terms of halves. Part a needs one logarithm.
Hint 2/4
$q = Q_0e^{-t/\tau}$, so the time to fall to a fraction $f$ is $t = \tau\ln(1/f)$; in particular the half life is $\tau\ln 2 = 0.693\,\tau$.
Hint 3/4
Here the half life is $3.00\ \mathrm{s}$, so $3.00 = \tau\ln 2$.
Hint 4/4
The time constant is $4.33\ \mathrm{s}$, and a quarter is left after $6.00\ \mathrm{s}$.
the half life is the measurable quantity in a laboratory and the time constant is the one in the formula, so this conversion is the bridge between the two
Use the structure rather than the formula
$$\tfrac14 = \left(\tfrac12\right)^{2} \Rightarrow t = 2t_{1/2} = 6.00\ \mathrm{s}$$
equal intervals multiply by equal factors, so two half lives is a quarter regardless of the numbers, and no logarithm is needed
Independent check on the second answer through the exponential: $e^{-6.00/4.33} = e^{-1.386} = 0.250$, a quarter exactly, using the time constant from part a rather than the halving argument.
The half life is $\tau\ln 2$, and further halvings simply multiply.
⚠ Taking the time constant to be the half life
both are described as the characteristic time of the decay, and half is the more familiar fraction
right$$U_{\rm source} = C\varepsilon^{2} = 2U_C,\qquad \text{the other half heats the resistance}$$
8.7Ammeters and voltmeters: the meter is part of the circuit
Connecting a meter changes the circuit you meant to measure, by an amount set by the meter's own resistance.
Every number on this page has been calculated, and in a laboratory it would be measured instead, which raises a question the calculations never had to face: does the act of measuring disturb the thing measured?
RuleResult 8.7: what each meter must be, and how far off it is
Conditions
The ammeter is inserted into a branch, so its resistance $R_A$ adds to that branch
The voltmeter is connected across an element, so its resistance $R_V$ is in parallel with that element
The circuit is otherwise ohmic and in a steady state
$$\boxed{\,\frac{I_{\rm read}}{I_{\rm true}} = \frac{R_{\rm loop}}{R_{\rm loop}+R_A},\qquad R \to \frac{RR_V}{R+R_V}\,}$$
An ammeter has to be cut into the path, so it adds resistance and always reads a little low; the smaller its own resistance compared with the rest of the loop, the smaller the error. A voltmeter is hung across the element, so it opens a second path and effectively lowers that element's resistance; the larger its own resistance compared with the element, the smaller the error.
Proof
The ammeter case is a single loop: the true current is $\varepsilon/R_{\rm loop}$ and the current once the meter is in the loop is $\varepsilon/(R_{\rm loop}+R_A)$. Dividing the second by the first gives the ratio in the box.
The voltmeter case is a parallel combination: the meter and the element are now both between the same two points, so what the rest of the circuit sees is the pair combined, $RR_V/(R+R_V)$, which is smaller than $R$.
A smaller resistance there takes a smaller share of the applied voltage, so the reading is below the value the circuit had before the meter arrived. The reading is honest about the circuit that now exists; it is the old circuit that has been disturbed.
In both cases the error vanishes in the same limit: the meter's resistance must be negligible compared with what is in series with it, and enormous compared with what is in parallel with it. Those are opposite requirements, which is why the two instruments are not interchangeable even when one box does both jobs.
Where each meter goes, and what that forces its resistance to be. The $\textcolor{#1f6feb}{\text{ammeter}}$ is cut into the path so that the same current passes through it, which is why it must add almost no resistance. The $\textcolor{#cf222e}{\text{voltmeter}}$ is hung across the element so that it shares the same two points, which is why it must draw almost no current.
Looks like this, but is not
This meter has an input resistance of $10.0\ \mathrm{M\Omega}$, which is enormous, so its readings can be taken as exact. Ten megohms really is enormous compared with the resistors in most laboratory circuits, where the loading error is genuinely negligible.
Enormous compared with what is the question the sentence forgets to ask. Put that meter across a $10.0\ \mathrm{M\Omega}$ resistor and the two are equal, the pair behaves as $5.00\ \mathrm{M\Omega}$, and the reading can be a long way from the undisturbed value. A meter's resistance is never large or small on its own; it is large or small compared with the resistance it is placed next to. The same sentence applied to an ammeter would be the opposite error: a $0.100\ \Omega$ ammeter is negligible in a $100\ \Omega$ loop and ruinous in a $0.200\ \Omega$ one.
A voltmeter that reads four volts where there were six
Two $10.0\ \mathrm{k\Omega}$ resistors in series are connected across an ideal $12.0\ \mathrm{V}$ source. A voltmeter is connected across the lower resistor. Find the reading if the voltmeter's resistance is $10.0\ \mathrm{k\Omega}$, and again if it is $1.00\ \mathrm{M\Omega}$. Compare both with the undisturbed value.
Given
two $10.0\ \mathrm{k\Omega}$ resistors in series across an ideal $12.0\ \mathrm{V}$ source
a voltmeter across the lower resistor
voltmeter resistance $10.0\ \mathrm{k\Omega}$ in the first case and $1.00\ \mathrm{M\Omega}$ in the second
Find
the two readings and the error in each
SolutionThe value there was before the meter arrived
the divider is recalculated with the new lower arm; the reading is honest about the circuit that now exists, and that circuit is not the one being studied
Independent check on the first reading through the currents rather than the divider: with the poor meter the total resistance is $15.0\ \mathrm{k\Omega}$, so the source drives $0.800\ \mathrm{mA}$, which produces $(0.800)(5.00) = 4.00\ \mathrm{V}$ across the parallel pair, as found.
The rule of thumb that follows: a voltmeter disturbs nothing if its resistance is at least a hundred times the resistance it is placed across. In circuits built from megohms, which is where this bites hardest, even a good meter is not automatically safe.
How small must an ammeter be to stay out of the way?
An ideal $12.0\ \mathrm{V}$ source drives a $4.00\ \Omega$ resistor. An ammeter of resistance $0.500\ \Omega$ is inserted to measure the current. Find the true current, the reading, and the largest ammeter resistance that would keep the reading within one per cent of the true value.
Given
ideal source, $\varepsilon = 12.0\ \mathrm{V}$
$R = 4.00\ \Omega$ alone in the loop
ammeter resistance $R_A = 0.500\ \Omega$, inserted in series
Find
the true current, the reading, and the tolerable meter resistance
the inequality is rearranged symbolically before any number goes in, so the direction of the inequality is decided once and cannot flip during the arithmetic
$$R_A \le (4.00)(0.0101) = 0.0404\ \Omega$$
about a hundredth of the loop resistance, which is the same factor of a hundred that the voltmeter needed, in the opposite direction
Independent check on the bound by putting it back in: with $R_A = 0.0404\ \Omega$ the loop is $4.0404\ \Omega$ and the current is $2.970\ \mathrm{A}$, which is $99.0\%$ of $3.00\ \mathrm{A}$ exactly as required.
Both meters obey the same sentence with the comparison reversed: a hundred times smaller than the loop for an ammeter, a hundred times larger than the element for a voltmeter. That is why an ammeter connected across a battery is a short circuit with a fuse in it, and why a voltmeter placed in series with a lamp leaves the lamp dark.
Checkpoint
§08.7 — a meter in the wrong place●●○○○
Thirty seconds, and the reasoning matters more than the verdict. A student wants to know the current in a lamp of resistance $12.0\ \Omega$ running from a $6.00\ \mathrm{V}$ source, and connects a voltmeter of resistance $1.00\ \mathrm{M\Omega}$ in series with the lamp.
voltmeter of resistance $1.00\ \mathrm{M\Omega}$ placed in series with the lamp
Find
(a) True or false: the lamp will glow roughly as brightly as before and the meter will show the voltage across the lamp. Give your reason in one sentence.
Hint 1/4
Work out what the loop resistance has become, and compare it with what it was. The lamp's brightness follows from the current, not from anybody's intention.
Hint 2/4
Elements in series share the same current, and the current in the loop is the emf divided by the total resistance in it.
Hint 3/4
Here the loop is now $12.0\ \Omega$ plus $1.00\times10^{6}\ \Omega$, driven by $6.00\ \mathrm{V}$.
Hint 4/4
False: the current collapses to about six microamperes and the lamp does not light at all.
Independent check by the loop rule: the two voltages must add to the emf, and $72\ \mathrm{\mu V} + 5.99993\ \mathrm{V} = 6.00\ \mathrm{V}$, so the meter really does take essentially all of it.
A misplaced meter reads honestly a circuit that it created itself.
⚠ Adding a voltmeter's resistance to the element it is measuring
the meter is a resistance and the element is a resistance, and adding is the first thing one does with two resistances
wrong$$R_{\rm eff} = R + R_V$$
right$$R_{\rm eff} = \frac{RR_V}{R+R_V} < R$$
⚠ Assuming a reading is the undisturbed value
the number on the display is a measurement, and measurements are treated as facts about the circuit that was there before
wrong$$V_{\rm true} = V_{\rm read}$$
right$$V_{\rm read} = V_{\rm true}\ \text{only in the limit } R_V \gg R$$
⚠ Connecting an ammeter across an element instead of into the path
connecting two probes across something is the habit built by using a voltmeter, and the two instruments often live in the same box
right$$\text{break the wire and put the ammeter in the gap, in series}$$
Reducing a network and coming back out again
Any circuit made only of sources and resistors in which some pair is genuinely in series or in parallel. Six steps, three inwards and three outwards, and the order never changes.
Redraw before calculating
Trace both ends of every resistor and mark what else is attached there. Two resistors are in series only if the same current has nowhere else to go between them, and in parallel only if both of their ends are joined to the same two conductors. Being drawn next to each other means nothing. If no pair passes either test, stop: this is a job for the junction and loop rules.
Collapse the innermost group first
Add the resistances for a series group, add the reciprocals and invert for a parallel group. Write the collapsed value onto the picture; you will need it on the way out.
Include the internal resistance and repeat
When the external network is one number, add the source's internal resistance to it. That total is $R_{\rm eq}$ as the source sees it, and nothing inside the network has been solved yet.
One division gives the current that everything hangs on
$I = \varepsilon/R_{\rm eq}$. This is the current in the source, in $r$, and in every element that lies on the single road before the network first splits. Get the terminal voltage at the same time, $V_{ab} = \varepsilon - Ir$.
Expand outwards, carrying the shared quantity
At each stage ask what the grouping forces its members to share: a series group shares the current, a parallel group shares the voltage. Hand every member the shared quantity, then use Ohm's law on each member alone to get the other one.
Check with sums you did not use
The voltages along any series path must add to the voltage across that path, the currents into any parallel group must add to the current that arrived, and the power balance $\varepsilon I = \sum I_j^{2}R_j$ must hold. All three use arithmetic that was not part of the solution, so all three are real tests.
Where it goes wrong
Leaving the internal resistance out of the equivalent resistance, which makes every current slightly too large.
Reporting a sum of reciprocals as a resistance, having forgotten the final inversion.
Using the total current in $I^{2}R$ for a resistor that is inside a parallel branch.
Kirchhoff without sign errors
Any circuit with more than one source, or one in which no pair of resistors is in series or in parallel. It always works, including on circuits that would reduce, so it is the fallback when you cannot see the structure.
Draw an arrow on every branch and name it
One current per branch, direction guessed. Do not try to guess well: a wrong guess costs nothing and produces a negative number at the end, which is a legitimate answer.
Count before writing
With $b$ branch currents you need $b$ independent equations. Take $j-1$ of them from the junctions, where $j$ is the number of junctions, and the remaining $b-j+1$ from loops.
Junction equations: everything in equals everything out
At each chosen junction, add the currents whose arrows point in and set them equal to the sum of those pointing out. Skip one junction entirely; its equation is the sum of all the others and carries nothing new.
Choose loops greedily
Pick a loop, walk it in a fixed direction, then pick the next loop so that it contains at least one branch none of the earlier loops used. Stop when you have enough. A loop whose every branch has already appeared will give you $0 = 0$.
Walk each loop applying the four signs mechanically
Through a resistance with the arrow, $-IR$; against the arrow, $+IR$. Through a source from its negative plate to its positive plate, $+\varepsilon$; the other way, $-\varepsilon$, whatever the current is doing. Do not think about which way the current really flows while you are walking; that is what the algebra is for.
Solve, then read the signs and check the energy
A negative current means that branch runs the other way; keep every equation as written and report the direction in words. Then check with the power balance, remembering that a source carrying current into its positive terminal is absorbing $\varepsilon I$ rather than delivering it.
Where it goes wrong
Writing $-IR$ for every resistor regardless of the direction of travel.
Adding the outer loop as an extra equation and then hunting for an algebra error that is not there.
Redrawing an arrow halfway through the solution, which invalidates every equation already written.
Any RC question, in five questions asked in this order
Anything with a capacitor and a switch. The five questions are always the same and four of them are answered before the time in the problem is even looked at.
Which resistance is the charge actually moving through?
Not the resistor nearest the capacitor on the page, but the total resistance of the loop the current is about to flow round. This is the resistance that goes into $\tau$, and it is different for the charging loop and the discharging loop in most real circuits.
What is $\tau = RC$?
Put both quantities into ohms and farads before multiplying. Kilohms with microfarads gives milliseconds and megohms with microfarads gives seconds, and mixing prefixes is the standard way to be wrong by a factor of a thousand.
What is the value at the very first instant?
An uncharged capacitor has no voltage across it, so at $t=0$ it behaves as a plain wire: the current is set by the resistance alone. A charged one behaves as a source of $q/C$ volts.
What is the value after a very long time?
The current in a capacitor branch is zero, so that branch behaves as a break: no current through it and therefore no voltage across any resistor in series with it. Solve the circuit that remains, then walk from the capacitor to a known point to get its voltage.
Only now put the time in
Every quantity moves from its initial value to its final value with the factor $e^{-t/\tau}$. A quantity that decays to zero carries the bare exponential; a quantity that grows to a final value carries $1-e^{-t/\tau}$. If a time is wanted instead, isolate the exponential first and take one logarithm.
Where it goes wrong
Using the charging resistance for a discharge that happens through something else.
Attaching the growth bracket to the current, which decays, or the bare exponential to the charge, which grows.
Treating five time constants as exactly full rather than as $99.3\%$ full, in a question that asks for a time.
Two lamps end to end on a tired battery
A battery with $\varepsilon = 12.0\ \mathrm{V}$ and $r = 1.00\ \Omega$ drives two lamps, each of resistance $6.00\ \Omega$, connected one after the other in a single loop. Find the current, the terminal voltage, and the power in each lamp.
Independent check by the power balance: the battery delivers $(12.0)(0.9231) = 11.1\ \mathrm{W}$, and the two lamps take $5.11$ each while the internal resistance takes $(0.9231)^{2}(1.00) = 0.852\ \mathrm{W}$, adding to $11.1\ \mathrm{W}$.
The same two lamps side by side on the same tired battery
The same battery, $\varepsilon = 12.0\ \mathrm{V}$ and $r = 1.00\ \Omega$, now drives the same two $6.00\ \Omega$ lamps connected across the same two terminals as each other. Find the current from the battery, the terminal voltage, and the power in each lamp.
Independent check by the power balance: the battery delivers $(12.0)(3.00) = 36.0\ \mathrm{W}$, the two lamps take $13.5$ each and the internal resistance takes $(3.00)^{2}(1.00) = 9.00\ \mathrm{W}$, adding to $36.0\ \mathrm{W}$. Note how much larger the wasted share has become.
Identical battery, identical lamps, and the only difference is which quantity the pair is forced to share. Side by side, each lamp is $2.6$ times brighter, but the battery is working three times harder and its terminal voltage has sagged from $11.1\ \mathrm{V}$ to $9.00\ \mathrm{V}$; the fraction of the emf wasted inside has gone from under eight per cent to a quarter. With an ideal battery the parallel lamps would each get the full $12.0\ \mathrm{V}$ and $24.0\ \mathrm{W}$, so almost half of the expected brightness has been lost to the internal resistance alone.
How to tell them apart
Ask one question before writing anything: do the two elements have the same current forced through them, or the same voltage forced across them? Series forces the current and the voltages divide; parallel forces the voltage and the currents add. Then ask a second question that only matters for a real source: how much current is the arrangement pulling in total? The parallel case pulls more, so the internal resistance takes a bigger cut, and that is why the two lamps do not simply behave as they would on their own.
Scaffolding comes off
The common skeleton
Name what is being asked for, and mark on the diagram which elements lie on a single road and which sit in parallel groups.
Reduce the external network to one resistance, innermost group first, then add the internal resistance of the source.
Get the current from the source with one division, and the terminal voltage with one subtraction.
Expand back outwards, handing each group the quantity it shares: the current along a single road, the voltage across a parallel group.
Apply Ohm's law to each individual element to get whichever of its current and voltage is still missing.
Compute any powers last, each from that element's own current or its own voltage.
Check with something you did not use: branch currents adding, voltages around a loop adding, or the power balance.
1 · fully worked
Eighteen volts through a network, every quantity found
A source of emf $18.0\ \mathrm{V}$ and internal resistance $1.00\ \Omega$ drives a $5.00\ \Omega$ resistor in series with a parallel combination of $4.00\ \Omega$ and $12.0\ \Omega$. Find every current, every voltage and every power in the circuit.
Independent check by the power balance: the source delivers $(18.0)(2.00) = 36.0\ \mathrm{W}$ and the four dissipations add to $4.00+20.0+9.00+3.00 = 36.0\ \mathrm{W}$. Second check: the branch currents add, $1.50+0.500 = 2.00\ \mathrm{A}$.
2 · you write the reasoning
Easier than rung 1 on purpose: one loop, no parallel group, three lines. A source of emf $9.00\ \mathrm{V}$ and internal resistance $0.500\ \Omega$ is connected to a single $4.00\ \Omega$ resistor. The three lines below are all correct. Before opening the model reasons, say in your own words why each line is allowed, and in particular what the third line is testing that the first two are not.
reasoning
The two resistances may be added only because the same current passes through both of them; that is what one loop with no junctions in it means. The internal resistance is included not as a special case but because the current in it is the same current, so it belongs in the same sum. If the internal resistance had been left out, the current would have come out as $2.25\ \mathrm{A}$, which is twelve per cent high, and nothing later in the calculation would have looked wrong.
reasoning
This line is the definition of terminal voltage applied with the current from the first line. The subtraction, rather than an addition, is because current is leaving the positive terminal of the source, so the internal resistance takes its share on the way out. The result is what a voltmeter across the terminals would show, and it is also exactly $(2.00)(4.00)$, the voltage across the external resistor, as it has to be since the external resistor is the only thing between those two terminals.
reasoning
This line is not a new result but a test. The left side is the rate at which the source converts chemical energy, the right side is the rate at which the two resistances heat up, and energy conservation demands they be equal. It counts as a genuine check because it combines both earlier lines in a way neither of them assumed: an error in the current would break it, and so would an error in either resistance. Recomputing the current a second way would not test anything, because it would use the same assumption.
3 · find the buried error
Harder than rung 2: two loops, two sources, and one of them turns out to be charging. Two nodes $a$ and $b$ are joined by three branches. Branch one has a $12.0\ \mathrm{V}$ source with $4.00\ \Omega$, branch two a $4.00\ \mathrm{V}$ source with $2.00\ \Omega$, and branch three a bare $4.00\ \Omega$ resistor. Both sources have their positive terminals towards $a$. The arrows are drawn with $I_1$ and $I_2$ running from $b$ to $a$ and $I_3$ running from $a$ to $b$. A student produces the four steps below and reports $I_1 = 1.75\ \mathrm{A}$, $I_2 = 0.500\ \mathrm{A}$ and $I_3 = 1.25\ \mathrm{A}$, all in the directions drawn. Exactly two of the four steps are faulty. Find them.
the two buried errors (2)
⚠ step 1
The junction equation contradicts the arrows in the problem. Both $I_1$ and $I_2$ are drawn arriving at node $a$ and only $I_3$ is drawn leaving it, so the rule reads $I_1+I_2 = I_3$, not $I_1 = I_2+I_3$. The version written would be correct for a different picture, one in which $I_2$ leaves the node.
One branch usually does carry the largest current and does appear to split, and the phrase the current splits gets attached to the branch with the biggest source in it rather than to the arrows actually drawn. The written equation also looks like the familiar form met in every simple parallel circuit, so nothing about it reads as odd.
right
Read the arrows, not the circuit. Everything whose arrow points into the node goes on one side, everything whose arrow points out goes on the other: $I_1+I_2 = I_3$. Whether an arrow was guessed correctly is irrelevant here and is settled later by the sign of the answer.
⚠ step 3
The sign of the resistor term is wrong. The walk goes from $b$ up through branch two, which is the direction in which $I_2$ was drawn, so crossing that resistor gives $-2.00I_2$ and the equation is $4.00-2.00I_2-4.00I_3 = 0$. As written, the term was given the sign that belongs to a walk against the arrow.
The two resistor terms in the same equation get opposite treatment here only because the walk happens to run with one arrow and with the other, and it is easy to copy the sign pattern of the previous loop equation instead of re-deriving it. It is also the kind of error that leaves the equation looking perfectly ordinary.
right
Apply the four cases mechanically and never by memory of the last equation you wrote: with the arrow, $-IR$; against it, $+IR$. Correcting both faults gives $I_1 = 1.75\ \mathrm{A}$, $I_3 = 1.25\ \mathrm{A}$ and $I_2 = -0.500\ \mathrm{A}$, so the second source is being charged at $0.500\ \mathrm{A}$ rather than delivering current, which is the opposite of what was reported.
4 · the bare problem
§08.4 — a two loop network with no scaffolding●●●●○
No hints on the page beyond the ladder you have just climbed, and the same skeleton applies. Two nodes $a$ and $b$ are joined by three branches. Branch one contains a $6.00\ \mathrm{V}$ source in series with $2.00\ \Omega$, branch two a $12.0\ \mathrm{V}$ source in series with $2.00\ \Omega$, and branch three a bare $4.00\ \Omega$ resistor. Both sources have their positive terminals facing node $a$.
Given
branch 1: $6.00\ \mathrm{V}$ with $2.00\ \Omega$, positive terminal towards $a$
branch 2: $12.0\ \mathrm{V}$ with $2.00\ \Omega$, positive terminal towards $a$
branch 3: $4.00\ \Omega$ alone
Find
(a) Find the current in each branch, stating its true direction.
(b) Find the potential difference between $a$ and $b$.
(c) Say which source, if any, is being charged, and check your answer with the power balance.
Hint 1/4
Three unknown currents means three independent equations. Draw and name the arrows first, then count: two junctions give one useful junction equation, so two loop equations are needed.
Hint 2/4
The junction rule is that arrows in equal arrows out. The loop rule is a walk applying four signs: $-IR$ with the arrow, $+IR$ against it, $+\varepsilon$ from the negative plate to the positive one, $-\varepsilon$ the other way.
Hint 3/4
With $I_1$ and $I_2$ drawn from $b$ to $a$ and $I_3$ from $a$ to $b$, the branches are $6.00\ \mathrm{V}$ with $2.00\ \Omega$, $12.0\ \mathrm{V}$ with $2.00\ \Omega$, and a bare $4.00\ \Omega$.
Hint 4/4
The currents are $I_1 = -0.600\ \mathrm{A}$, $I_2 = 2.40\ \mathrm{A}$ and $I_3 = 1.80\ \mathrm{A}$, and $V_{ab} = 7.20\ \mathrm{V}$.
Show solutionLabel and count
$$I_1+I_2 = I_3$$
the junction rule at node $a$ with two arrows in and one out; the equation at $b$ would repeat it
Independent check by computing $V_{ab}$ along the other two branches, which the solution did not use: through branch one, $6.00-(2.00)(-0.600) = 7.20\ \mathrm{V}$; through branch two, $12.0-(2.00)(2.40) = 7.20\ \mathrm{V}$. Three routes, one number. The power balance also closes at $28.8\ \mathrm{W}$ on both sides.
Solve each loop for its own current; a minus means direction.
Full exam-style question
Full exam question: a network with internal resistance and a capacitor at steady stateexam format
A battery of emf $24.0\ \mathrm{V}$ and internal resistance $1.00\ \Omega$ is connected to the following network. A $3.00\ \Omega$ resistor is in series with a parallel combination of $6.00\ \Omega$ and $12.0\ \Omega$, and that group is in series with a $2.00\ \Omega$ resistor. A $10.0\ \mathrm{\mu F}$ capacitor is connected across the $2.00\ \Omega$ resistor. The circuit has been switched on for a long time. Find (a) the current from the battery and the terminal voltage, (b) the current in each of the two parallel resistors, (c) the charge and the stored energy on the capacitor, and (d) verify your solution with the power balance.
Two independent checks, neither used in the solution. First, the external drops must add to the terminal voltage: $7.20+9.60+4.80 = 21.6\ \mathrm{V}$. Second, the branch currents must add to the total: $1.60+0.800 = 2.40\ \mathrm{A}$. The energy stored is a tenth of a millijoule, which is the right order for ten microfarads at a few volts.
Four parts, one reduction, one division and eight lines of expansion. In an examination the capacitor line in part c is worth as much as the whole of part b and takes ten seconds, provided the steady state statement is made first.
The shape of this question is the one to expect: a network that reduces, a real source so that the terminal voltage is not the emf, and a capacitor placed so that it is doing nothing except reporting the voltage of the element it sits across. The capacitor never enters the resistor arithmetic at all.
Practice
A · concept 4 questions
1§08.2 — does a new branch steal current from the old one?●●●○○
One mark, and the reasoning is the mark. A resistor is connected across an ideal source and carries a steady current. A second resistor is then connected in parallel with the first, across the same two terminals.
Given
the first resistor carries a steady current before the change
the source is ideal, so its terminal voltage cannot sag
a second resistor is added in parallel with the first
Find
(a) True or false: the current in the first resistor falls, because the current now has to be shared between two resistors. Give your reason in one sentence.
Hint 1/4
Write down the two quantities that fix the current in the first resistor, and then ask whether the second resistor changed either of them.
Hint 2/4
The current in a branch is the potential difference across that branch divided by its own resistance, and the terminal voltage of an ideal source does not depend on what is drawn from it.
Hint 3/4
Here the branch voltage is still the source voltage and the branch resistance is untouched, because the new resistor was added across the same two terminals rather than into the same path.
Hint 4/4
False: the first current is unchanged and it is the total current from the source that rises.
Show solutionIdentify what the branch current depends on
$$I_1 = \frac{V}{R_1}$$
only two quantities appear, so only a change in one of them can change the current
Independent check with numbers: $12.0\ \mathrm{V}$ across $4.00\ \Omega$ gives $3.00\ \mathrm{A}$; adding a $6.00\ \Omega$ branch gives $R_{\rm par} = 2.40\ \Omega$ and a total current of $5.00\ \mathrm{A}$, of which the first branch still takes $12.0/4.00 = 3.00\ \mathrm{A}$.
A new parallel branch raises the total, not the old branch.
2§08.3 — one lamp fails and the others react●●●○○
One mark, and it is the classic. Three identical lamps are connected to an ideal source: lamp L1 is in series with a parallel pair made of lamps L2 and L3. All three have the same resistance. Lamp L3 then burns out, which leaves its branch open.
Given
three identical lamps of equal resistance
L1 in series with the parallel pair L2 and L3
an ideal source, and L3 becomes an
Find
(a) What happens to the brightness of L1 and of L2?
Hint 1/4
Two separate questions hide in this one. L1 carries the total current, so it follows the equivalent resistance. L2 has its own branch, so it follows the voltage across the parallel group.
Hint 2/4
For identical lamps of resistance $R$: a parallel pair is $R/2$, and the equivalent resistance of the whole is that plus $R$. Brightness follows the power in each lamp, $I^{2}R$ with that lamp's own current.
Hint 3/4
Before: $R_{\rm eq} = R+R/2 = 1.5R$, so the total current is $V/1.5R$ and L2 takes half of it. After: $R_{\rm eq} = 2R$, so the total is $V/2R$ and L2 takes all of it.
Hint 4/4
L1 falls from $0.67V/R$ to $0.50V/R$, and L2 rises from $0.33V/R$ to $0.50V/R$.
Independent check by the total power, which was not used: before, $P_{\rm tot} = V^{2}/1.5R = 0.667V^{2}/R$; after, $V^{2}/2R = 0.500V^{2}/R$. Less total power from the source, yet one lamp is brighter, which is only possible because the share has been redistributed.
Delete the open branch, recompute, and square only at the end.
3§08.1 — can a terminal voltage exceed the emf?●●●○○
One mark. A student measures $6.20\ \mathrm{V}$ across the terminals of a cell whose label reads $6.00\ \mathrm{V}$, while a current of $1.00\ \mathrm{A}$ passes through it, and concludes that either the meter or the label must be faulty.
Given
the cell is labelled $6.00\ \mathrm{V}$
a voltmeter across its terminals reads $6.20\ \mathrm{V}$
a current of $1.00\ \mathrm{A}$ is passing through the cell
Find
(a) True or false: a terminal voltage above the emf is impossible, so something must be faulty. Give your reason in one sentence.
Hint 1/4
The formula for terminal voltage has a sign in it that depends on which way the current is going through the source. Ask what happens when the current is pushed in at the positive terminal instead of drawn out of it.
Hint 2/4
$V_{ab} = \varepsilon - Ir$ when the source is delivering current, and $V_{ab} = \varepsilon + Ir$ when current is being driven into its positive terminal by something else in the circuit.
Hint 3/4
With $\varepsilon = 6.00\ \mathrm{V}$, $I = 1.00\ \mathrm{A}$ and a reading of $6.20\ \mathrm{V}$, the excess of $0.20\ \mathrm{V}$ is $Ir$ with $r = 0.200\ \Omega$.
Hint 4/4
False: this is exactly what a cell being charged looks like, and it puts the internal resistance at $0.200\ \Omega$.
Show solutionChoose the right sign
$$V_{ab} = \varepsilon + Ir\ \text{when current enters the positive terminal}$$
the internal resistance always opposes the current, so whether its drop is taken off or added on depends on which way the current is being driven
the excess over the emf is exactly the internal drop, so the anomalous reading is the measurement of $r$
Answer $$\boxed{\,\text{False: the cell is being charged, and } r = 0.200\ \Omega\,}$$
Check
Independent check by energy: the cell is receiving $\varepsilon I = 6.00\ \mathrm{W}$ chemically and dissipating $I^{2}r = 0.200\ \mathrm{W}$ internally, so whatever is charging it must be supplying $6.20\ \mathrm{W}$ at its terminals, which is $V_{ab}I$ with the measured reading.
A terminal voltage above the emf means charging, not a fault.
4§08.6 — the energy books for a charging capacitor●●●○○
One mark, and it is the result people find hardest to believe. An uncharged capacitor is charged to a final voltage equal to the emf, from a constant source, through a resistor.
Given
the capacitor starts empty and ends at the emf of the source
the source has a constant emf
all the charge passes through one resistor on the way
Find
(a) Which statement about the energy is correct?
Hint 1/4
Three energies are involved and only two of them are worth computing: what leaves the source, and what stays in the capacitor. The third is the difference.
Hint 2/4
Every coulomb that leaves the source crosses the full emf, so the source delivers $Q_f\varepsilon = C\varepsilon^{2}$. The capacitor ends up holding $\tfrac12 C\varepsilon^{2}$.
Hint 3/4
The capacitor is charged to the emf and the source is constant, so the two expressions above are the whole of the accounting; the resistance does not appear in either of them.
Hint 4/4
The difference is $\tfrac12 C\varepsilon^{2}$ of heat, exactly equal to what is stored, whatever the resistance is.
the capacitor's own voltage rose from zero to $\varepsilon$, which is where its factor of one half comes from, and energy conservation supplies the rest
Independent check by integrating the heating rather than subtracting: $\int_0^{\infty}I^{2}R\,dt$ with $I = (\varepsilon/R)e^{-t/RC}$ gives $(\varepsilon^{2}/R)(RC/2) = \tfrac12 C\varepsilon^{2}$, and the resistance cancels between the height of the curve and its width.
Charging through any resistance stores half and wastes half.
B · computation 8 questions
1§08.1 — measuring an emf and an internal resistance●●○○○
A battery is tested twice with different loads and a voltmeter across its terminals. With the first load it delivers $2.00\ \mathrm{A}$ and the terminals read $11.6\ \mathrm{V}$; with the second it delivers $5.00\ \mathrm{A}$ and the terminals read $11.0\ \mathrm{V}$.
Given
$I = 2.00\ \mathrm{A}$ with $V_{ab} = 11.6\ \mathrm{V}$
$I = 5.00\ \mathrm{A}$ with $V_{ab} = 11.0\ \mathrm{V}$
the emf and the internal resistance are the same in both tests
Find
(a) Find the emf and the internal resistance.
(b) Find the current if the terminals were joined by a wire of negligible resistance.
(c) Find the external resistance used in the first test.
Hint 1/4
Two unknowns and two measurements: this is a pair of simultaneous equations, and subtracting one from the other removes the emf in a single line.
Hint 2/4
$V_{ab} = \varepsilon - Ir$ for a source delivering current, and the largest current a source can give is $\varepsilon/r$.
Hint 3/4
The two measurements give $11.6 = \varepsilon - 2.00r$ and $11.0 = \varepsilon - 5.00r$.
Hint 4/4
Subtracting gives $r = 0.200\ \Omega$ and $\varepsilon = 12.0\ \mathrm{V}$, so the short circuit current would be $60.0\ \mathrm{A}$ and the first load was $5.80\ \Omega$.
the external resistor has the terminal voltage across it, not the emf, which is the whole point of the exercise
Answer $$\boxed{\,\varepsilon = 12.0\ \mathrm{V},\ r = 0.200\ \Omega,\ I_{\rm max} = 60.0\ \mathrm{A},\ R = 5.80\ \Omega\,}$$
Check
Independent check of the second measurement, which was used only in the subtraction: $\varepsilon - Ir = 12.0-(5.00)(0.200) = 11.0\ \mathrm{V}$, exactly the reading given. Order of magnitude: a car battery with two tenths of an ohm and sixty amperes on short circuit is realistic for a small or ageing one.
Two loads and two readings; subtract to eliminate the emf first.
2§08.2 — a four resistor network reduced to one number●●●○○
A network is built as follows. A $6.00\ \Omega$ and a $3.00\ \Omega$ resistor are in parallel with each other; that pair is in series with a $4.00\ \Omega$ resistor; and the whole of that is in parallel with a $12.0\ \Omega$ resistor. The combination is connected across an ideal $24.0\ \mathrm{V}$ source.
Given
$6.00\ \Omega$ in parallel with $3.00\ \Omega$
that pair in series with $4.00\ \Omega$
all of that in parallel with $12.0\ \Omega$, across an ideal $24.0\ \mathrm{V}$ source
Find
(a) Find the equivalent resistance of the network.
(b) Find the total current drawn from the source.
(c) Find the current in the $3.00\ \Omega$ resistor.
Hint 1/4
Work from the inside out for part a, then from the outside in for part c. The two journeys use different shared quantities and it is worth naming which is which before starting.
Hint 2/4
Product over sum for a parallel pair, plain addition for a series group. On the way back, a parallel group shares its voltage and a series group shares its current.
Hint 3/4
Inwards: $6.00$ with $3.00$ gives $2.00\ \Omega$; plus $4.00$ gives $6.00\ \Omega$; that with $12.0$ gives the answer to part a. The source holds $24.0\ \mathrm{V}$ across the whole thing.
Hint 4/4
$R_{\rm eq} = 4.00\ \Omega$, the total current is $6.00\ \mathrm{A}$, and the $3.00\ \Omega$ resistor carries $2.67\ \mathrm{A}$.
Independent check by adding the currents that were never added: the twelve ohm branch carries $24.0/12.0 = 2.00\ \mathrm{A}$ and the arm carries $4.00\ \mathrm{A}$, giving $6.00\ \mathrm{A}$ in total; and inside the arm, $8.00/6.00 = 1.33\ \mathrm{A}$ plus $2.67\ \mathrm{A}$ is $4.00\ \mathrm{A}$.
Reduce innermost first, expand last, peeling drops rather than splitting totals.
3§08.3 — a complete network with a real source●●●○○
A source of emf $15.0\ \mathrm{V}$ and internal resistance $0.500\ \Omega$ drives a $2.50\ \Omega$ resistor in series with a parallel combination of $6.00\ \Omega$ and $3.00\ \Omega$.
$R_2 = 6.00\ \Omega$ in parallel with $R_3 = 3.00\ \Omega$
Find
(a) Find the current from the source and the terminal voltage.
(b) Find the current in each of the parallel resistors.
(c) Find the power dissipated in each of the four resistances, and check the total.
Hint 1/4
One reduction inwards, one division, then a walk outwards. Decide before you start which elements carry the total current and which do not, and mark them.
Hint 2/4
$R_{\rm eq}$ includes $r$; then $I = \varepsilon/R_{\rm eq}$ and $V_{ab} = \varepsilon - Ir$. Each parallel branch obeys $I_i = V/R_i$ on its own, and each power uses that element's own current.
Hint 3/4
Here $6.00\parallel 3.00 = 2.00\ \Omega$, so $R_{\rm eq} = 0.500+2.50+2.00 = 5.00\ \Omega$ across $15.0\ \mathrm{V}$.
Hint 4/4
The current is $3.00\ \mathrm{A}$ with a terminal voltage of $13.5\ \mathrm{V}$; the branches carry $1.00\ \mathrm{A}$ and $2.00\ \mathrm{A}$; the powers are $4.50$, $22.5$, $6.00$ and $12.0\ \mathrm{W}$.
Independent check by the power balance: the source delivers $(15.0)(3.00) = 45.0\ \mathrm{W}$ and the four dissipations add to $4.50+22.5+6.00+12.0 = 45.0\ \mathrm{W}$. Second check: the branch currents add to $3.00\ \mathrm{A}$.
Internal resistance joins the series sum, and powers need their own currents.
4§08.4 — three branches between two nodes●●●●○
Two nodes $a$ and $b$ are joined by three branches. Branch one contains a $10.0\ \mathrm{V}$ source in series with $1.00\ \Omega$, branch two a $5.00\ \mathrm{V}$ source in series with $1.00\ \Omega$, and branch three a bare $2.00\ \Omega$ resistor. Both sources have their positive terminals towards $a$.
Given
branch 1: $10.0\ \mathrm{V}$ with $1.00\ \Omega$, positive terminal towards $a$
branch 2: $5.00\ \mathrm{V}$ with $1.00\ \Omega$, positive terminal towards $a$
branch 3: a bare $2.00\ \Omega$ resistor
Find
(a) Find the current in each branch and state its true direction.
(b) Find $V_{ab}$.
(c) State which source is delivering energy and which is receiving it, with the rate in each case.
Hint 1/4
Three unknown currents, so three independent equations: one junction equation and two loop equations. Draw and name the arrows before writing anything.
Hint 2/4
Junction: arrows in equal arrows out. Loop: $-IR$ walking with an arrow, $+IR$ against it, $+\varepsilon$ from the negative plate to the positive one.
Hint 3/4
With $I_1$ and $I_2$ drawn from $b$ to $a$ and $I_3$ from $a$ to $b$: $10.0-1.00I_1-2.00I_3 = 0$ and $5.00-1.00I_2-2.00I_3 = 0$, with $I_1+I_2 = I_3$.
Hint 4/4
The currents are $4.00\ \mathrm{A}$, $-1.00\ \mathrm{A}$ and $3.00\ \mathrm{A}$, and $V_{ab} = 6.00\ \mathrm{V}$.
Show solutionSet up
$$I_1+I_2 = I_3$$
the junction rule at $a$, with two arrows arriving and one leaving
$$I_1 = 10.0-2.00I_3,\qquad I_2 = 5.00-2.00I_3$$
each loop equation solved for its own branch current, so that the junction rule becomes one equation in one unknown
Independent check of $V_{ab}$ along the other two branches: $10.0-(1.00)(4.00) = 6.00\ \mathrm{V}$ and $5.00-(1.00)(-1.00) = 6.00\ \mathrm{V}$. Three routes, one number, and the power balance closes at $40.0\ \mathrm{W}$ on both sides.
Read the node voltage along the branch with no source.
5§08.1 — six cells, and one of them fitted backwards●●●○○
Six identical cells, each of emf $1.50\ \mathrm{V}$ and internal resistance $0.250\ \Omega$, are connected end to end to drive a $3.00\ \Omega$ lamp. Somebody then fits one of the six the wrong way round.
Given
six cells, each $\varepsilon = 1.50\ \mathrm{V}$ and $r = 0.250\ \Omega$
connected in series with a $3.00\ \Omega$ lamp
in the second case one cell is reversed
Find
(a) Find the current and the voltage across the lamp with all six correctly fitted.
(b) Find the current with one cell reversed.
(c) Find the rate at which the reversed cell is being charged.
Hint 1/4
A reversed cell changes the sum of the emfs but not the sum of the resistances. Deal with those two sums separately and the rest is a single loop.
Hint 2/4
In a series chain the emfs add with a sign that depends on their direction, while the internal resistances always add. Then $I = \varepsilon_{\rm tot}/(R+r_{\rm tot})$.
Hint 3/4
Six cells of $1.50\ \mathrm{V}$ and $0.250\ \Omega$ with a $3.00\ \Omega$ lamp: $r_{\rm tot} = 1.50\ \Omega$ in both cases, while $\varepsilon_{\rm tot}$ goes from $9.00\ \mathrm{V}$ to $9.00-2(1.50) = 6.00\ \mathrm{V}$.
Hint 4/4
The current falls from $2.00\ \mathrm{A}$ to $1.33\ \mathrm{A}$, and the reversed cell absorbs $2.00\ \mathrm{W}$.
Independent check on the second case by the power balance: the five good cells deliver $5(1.50)(1.333) = 10.0\ \mathrm{W}$; the reversed one takes $2.00\ \mathrm{W}$ chemically, the six internal resistances take $(1.333)^{2}(1.50) = 2.67\ \mathrm{W}$ and the lamp takes $(1.333)^{2}(3.00) = 5.33\ \mathrm{W}$, adding to $10.0\ \mathrm{W}$.
Emfs add with sign and resistances always; a reversed cell subtracts.
6§08.5 — a charging circuit read at three moments●●●○○
A $6.00\ \mathrm{V}$ source of negligible internal resistance, a $50.0\ \mathrm{k\Omega}$ resistor and an uncharged $4.00\ \mathrm{\mu F}$ capacitor are connected in one loop and the switch is closed at $t = 0$.
(a) Find the time constant, the final charge and the current at the instant of closing.
(b) Find the charge and the current at $t = 0.400\ \mathrm{s}$.
(c) Find the time at which the capacitor holds half its final charge.
Hint 1/4
Compute the three constants of the circuit before looking at any time in the question. Every later number is one of them multiplied by an exponential factor.
Hint 2/4
$\tau = RC$, $Q_f = C\varepsilon$, $I_0 = \varepsilon/R$; then $q = Q_f(1-e^{-t/\tau})$ and $I = I_0e^{-t/\tau}$.
Hint 3/4
With $R = 5.00\times10^{4}\ \Omega$, $C = 4.00\times10^{-6}\ \mathrm{F}$ and $\varepsilon = 6.00\ \mathrm{V}$, the time asked for in part b is $t/\tau = 2.00$.
Independent check that the loop rule holds at $t = 0.400\ \mathrm{s}$: the capacitor has $q/C = 20.8/4.00 = 5.19\ \mathrm{V}$ and the resistor has $IR = (1.62\times10^{-5})(5.00\times10^{4}) = 0.81\ \mathrm{V}$, and the two add to $6.00\ \mathrm{V}$.
Fix the pace and both endpoints before any time appears.
7§08.6 — a discharge, timed and accounted for●●●○○
A $100\ \mathrm{\mu F}$ capacitor is charged to $50.0\ \mathrm{V}$, disconnected from its source, and then connected across a $2.00\ \mathrm{k\Omega}$ resistor at $t = 0$.
(a) Find the time constant, the initial charge and the initial current.
(b) Find how long it takes for the voltage to fall to $10.0\ \mathrm{V}$.
(c) Find the total heat produced in the resistor over the whole discharge.
Hint 1/4
The capacitor is the source now, so start from its own voltage. For part c, ask where the stored energy can possibly go when there is nothing else in the loop.
Hint 2/4
$\tau = RC$, $I_0 = V_0/R$, and $V = V_0e^{-t/\tau}$, so the time to reach a fraction $f$ is $t = \tau\ln(1/f)$. The stored energy is $\tfrac12 CV_0^{2}$.
Hint 3/4
With $R = 2.00\times10^{3}\ \Omega$, $C = 1.00\times10^{-4}\ \mathrm{F}$ and $V_0 = 50.0\ \mathrm{V}$, part b asks for the time at which the fraction is $10.0/50.0$.
Hint 4/4
$\tau = 0.200\ \mathrm{s}$, $Q_0 = 5.00\ \mathrm{mC}$, $I_0 = 25.0\ \mathrm{mA}$; the voltage reaches $10.0\ \mathrm{V}$ at $0.322\ \mathrm{s}$; the total heat is $0.125\ \mathrm{J}$.
Independent check on the heat by integrating rather than by the stored energy: the initial heating rate is $I_0^{2}R = (0.0250)^{2}(2000) = 1.25\ \mathrm{W}$, decaying with time constant $\tau/2 = 0.100\ \mathrm{s}$, so the total is $(1.25)(0.100) = 0.125\ \mathrm{J}$.
The conducting loop owns the time constant, not the charging one.
8§08.7 — how much a voltmeter changes what it reads●●●●○
A $1.00\ \mathrm{k\Omega}$ and a $2.00\ \mathrm{k\Omega}$ resistor are connected in series across an ideal $12.0\ \mathrm{V}$ source. A voltmeter is then connected across the $2.00\ \mathrm{k\Omega}$ resistor.
Given
$1.00\ \mathrm{k\Omega}$ in series with $2.00\ \mathrm{k\Omega}$ across an ideal $12.0\ \mathrm{V}$ source
the voltmeter is connected across the $2.00\ \mathrm{k\Omega}$ resistor
voltmeter resistance $2.00\ \mathrm{k\Omega}$ in the first case
Find
(a) Find the potential difference across the $2.00\ \mathrm{k\Omega}$ resistor with no meter connected.
(b) Find the reading of a voltmeter whose own resistance is $2.00\ \mathrm{k\Omega}$.
(c) Find the smallest voltmeter resistance that would keep the reading within one per cent of the undisturbed value.
Hint 1/4
The meter is a resistor like any other. Redraw the circuit with it in place, and the question becomes an ordinary divider problem with a different lower arm.
Hint 2/4
A divider gives $V = \varepsilon R_{\rm lower}/(R_{\rm upper}+R_{\rm lower})$, and the meter in parallel with the lower arm replaces it by $RR_V/(R+R_V)$.
Hint 3/4
With $R_{\rm upper} = 1.00\ \mathrm{k\Omega}$, $R_{\rm lower} = 2.00\ \mathrm{k\Omega}$ and $\varepsilon = 12.0\ \mathrm{V}$: the undisturbed lower arm is $2.00\ \mathrm{k\Omega}$, and with a $2.00\ \mathrm{k\Omega}$ meter it becomes $1.00\ \mathrm{k\Omega}$.
Hint 4/4
The true value is $8.00\ \mathrm{V}$, the poor meter reads $6.00\ \mathrm{V}$, and a meter of at least $66.0\ \mathrm{k\Omega}$ is needed for one per cent.
solving the parallel expression for $R_V$; the answer is about thirty three times the resistance being measured, which is the usual order of the requirement
Independent check on the bound by substituting it back: $R_p = (2.00)(66.0)/(68.0) = 1.941\ \mathrm{k\Omega}$, and $12.0\times 1.941/2.941 = 7.92\ \mathrm{V}$, which is exactly $99.0\%$ of $8.00\ \mathrm{V}$.
A meter is a second path; bound the parallel arm first.
C · exam level 4 questions
1§08.3 — which resistor gets hottest●●●○○
Exam level, one mark, and it is decided before any arithmetic if you know what to compare. A $2.00\ \Omega$ resistor is in series with two $4.00\ \Omega$ resistors that are in parallel with each other, all across an ideal $12.0\ \mathrm{V}$ source.
Given
$R_1 = 2.00\ \Omega$ in series
two $4.00\ \Omega$ resistors in parallel with each other
ideal source of $12.0\ \mathrm{V}$
Find
(a) Which resistor dissipates the most power, and how much?
Hint 1/4
Every power needs that element's own current. Sort the three resistors into those carrying the total current and those carrying only part of it, before computing anything.
Hint 2/4
Identical resistors in parallel give $R/2$ and split the current evenly; series resistances add; and $P = I^{2}R$ with the current in that element.
Hint 3/4
Here $R_{\rm par} = 2.00\ \Omega$, so $R_{\rm eq} = 4.00\ \Omega$ and the total current is $12.0/4.00$, of which each parallel branch takes half.
Hint 4/4
The series resistor carries $3.00\ \mathrm{A}$ and dissipates $18.0\ \mathrm{W}$, while each parallel branch carries $1.50\ \mathrm{A}$ and dissipates $9.00\ \mathrm{W}$.
Independent check by the power balance: the source delivers $(12.0)(3.00) = 36.0\ \mathrm{W}$, and $18.0+9.00+9.00 = 36.0\ \mathrm{W}$.
Whatever lies on the single road carries everything and runs hottest.
2§08.3 — full network, real source, capacitor at steady state●●●●○
Exam level, four parts. A source of emf $21.0\ \mathrm{V}$ and internal resistance $0.500\ \Omega$ drives a $4.00\ \Omega$ resistor in series with a parallel combination of $10.0\ \Omega$ and $15.0\ \Omega$. A $5.00\ \mathrm{\mu F}$ capacitor is connected across the parallel combination, and the circuit has been running for a long time.
$R_1 = 4.00\ \Omega$ in series with $R_2 = 10.0\ \Omega$ parallel to $R_3 = 15.0\ \Omega$
$C = 5.00\ \mathrm{\mu F}$ across the parallel pair, steady state
Find
(a) Find the current from the source and the terminal voltage.
(b) Find the current in each of the parallel resistors.
(c) Find the charge and the energy on the capacitor.
(d) Check the whole solution with the power balance.
Hint 1/4
Settle what the capacitor is doing before touching the resistors: in a steady state its branch carries no current, so it takes no part in the network and only reports a voltage at the end.
Hint 2/4
$R_{\rm eq}$ includes $r$; then $I = \varepsilon/R_{\rm eq}$, $V_{ab} = \varepsilon-Ir$, each parallel branch takes $V/R_i$, and $q = CV_C$ with $U = \tfrac12 CV_C^{2}$.
Hint 3/4
Here $10.0\parallel 15.0 = 6.00\ \Omega$, so $R_{\rm eq} = 0.500+4.00+6.00 = 10.5\ \Omega$, driven by $21.0\ \mathrm{V}$, with the capacitor across the $6.00\ \Omega$ group.
Hint 4/4
The current is $2.00\ \mathrm{A}$ and the terminal voltage $20.0\ \mathrm{V}$; the branches carry $1.20\ \mathrm{A}$ and $0.800\ \mathrm{A}$; the capacitor holds $60.0\ \mathrm{\mu C}$ and $360\ \mathrm{\mu J}$.
Show solutionSet the capacitor aside
$$I_C = 0 \Rightarrow \text{the capacitor branch is a break in the steady state}$$
no charge is arriving on the plates, so the capacitor contributes nothing to the resistor network and only reports the voltage it sits across
Two independent checks. The external drops must add to the terminal voltage: $8.00+12.0 = 20.0\ \mathrm{V}$. The power balance: the source gives $(21.0)(2.00) = 42.0\ \mathrm{W}$, and $I^{2}r = 2.00$, $I^{2}R_1 = 16.0$, $V^{2}/R_2 = 14.4$ and $V^{2}/R_3 = 9.60\ \mathrm{W}$ add to $42.0\ \mathrm{W}$.
A steady state capacitor is a break that only reports a voltage.
3§08.4 — a branch that carries no current at all●●●●●
Exam level, three parts, and part a has an answer that looks like a mistake until you check it. Two nodes are joined by three branches: a $12.0\ \mathrm{V}$ source with internal resistance $1.00\ \Omega$, a $9.00\ \mathrm{V}$ source with internal resistance $1.00\ \Omega$, and a $3.00\ \Omega$ resistor. Both sources have their positive terminals towards the upper node.
branch 3: $R = 3.00\ \Omega$, and both sources point the same way
Find
(a) Find the current in each branch.
(b) Find the terminal voltage of each source and explain the value you get for the second one.
(c) The $3.00\ \Omega$ resistor is replaced by a $2.00\ \Omega$ one. Find the three currents again.
Hint 1/4
Set up the three equations in the usual way and let the algebra tell you what the second branch is doing. Do not decide in advance that every source must be carrying current.
Hint 2/4
Junction: arrows in equal arrows out. Loop: $-IR$ with an arrow, $+IR$ against it, $+\varepsilon$ from the negative plate to the positive one. Terminal voltage is $\varepsilon-Ir$ when the source delivers.
Hint 3/4
With $I_1$ and $I_2$ drawn towards the upper node and $I_3$ away from it: $12.0-1.00I_1-3.00I_3 = 0$, $9.00-1.00I_2-3.00I_3 = 0$, and $I_1+I_2 = I_3$.
Hint 4/4
The currents are $3.00\ \mathrm{A}$, zero and $3.00\ \mathrm{A}$; with the $2.00\ \Omega$ resistor they become $3.60\ \mathrm{A}$, $0.600\ \mathrm{A}$ and $4.20\ \mathrm{A}$.
with no current the second source loses nothing internally, so its terminals read its emf exactly, which is the only condition under which that happens
Independent check of the first case by computing the node voltage along all three branches: $12.0-(3.00)(1.00) = 9.00\ \mathrm{V}$, $9.00-(0)(1.00) = 9.00\ \mathrm{V}$ and $(3.00)(3.00) = 9.00\ \mathrm{V}$. Three routes, one number, and the second branch is consistent only because its current is zero.
Zero current is a real answer; terminals then read the emf.
4§08.5 — charged for one time constant, then dumped●●●●○
Exam level, three parts, and the second half uses a different resistance from the first. A $24.0\ \mathrm{V}$ source of negligible internal resistance charges an uncharged $2.00\ \mathrm{\mu F}$ capacitor through a $100\ \mathrm{k\Omega}$ resistor. After exactly $0.200\ \mathrm{s}$ the source is disconnected and the capacitor is immediately connected across a $25.0\ \mathrm{k\Omega}$ resistor instead.
Given
charging: $\varepsilon = 24.0\ \mathrm{V}$ through $R = 100\ \mathrm{k\Omega}$ into $C = 2.00\ \mathrm{\mu F}$, initially empty
the charging lasts exactly $0.200\ \mathrm{s}$
discharging: the same capacitor through $25.0\ \mathrm{k\Omega}$
Find
(a) Find the charge and the voltage on the capacitor at the moment the source is disconnected.
(b) Find the current at the first instant of the discharge.
(c) Find the charge remaining $0.100\ \mathrm{s}$ after the discharge begins.
Hint 1/4
Two circuits, one after the other, with the end of the first supplying the starting value of the second. Each has its own time constant, and they are not equal.
Hint 2/4
Charging: $q = C\varepsilon(1-e^{-t/\tau_c})$ with $\tau_c = R_cC$. Discharging: $q = Q_0e^{-t/\tau_d}$ with $\tau_d = R_dC$, and the initial discharge current is the capacitor's own voltage divided by the discharge resistance.
Hint 3/4
Here $\tau_c = (10^{5})(2.00\times10^{-6}) = 0.200\ \mathrm{s}$ so the charging lasts exactly one of them, and $\tau_d = (2.50\times10^{4})(2.00\times10^{-6}) = 0.0500\ \mathrm{s}$, so the discharge time asked for is two of those.
Hint 4/4
The capacitor reaches $30.3\ \mathrm{\mu C}$ at $15.2\ \mathrm{V}$, the discharge starts at $0.607\ \mathrm{mA}$, and $4.11\ \mathrm{\mu C}$ is left after two discharge time constants.
the discharge runs through a different resistance, so the charging time constant must not be carried over; this is the step the question is really testing
Independent check on part b through the charging current instead: at the end of the charging phase the current was $(24.0/10^{5})e^{-1} = 0.0883\ \mathrm{mA}$, and the discharge current starts about seven times larger because the resistance is four times smaller while the driving voltage is $15.2\ \mathrm{V}$ rather than the $8.8\ \mathrm{V}$ that was left across the charging resistor. The ratio $15.17/8.83 \times 4 = 6.9$ agrees with $0.607/0.0883 = 6.9$.
Rewiring brings a new time constant and a new starting value.
D · interleaved 4 questions
1§08.5 — two capacitors across a working battery●●●●○
This set is deliberately mixed, so decide for yourself which tool each question wants before reaching for one. A battery of emf $12.0\ \mathrm{V}$ and internal resistance $2.00\ \Omega$ drives a $10.0\ \Omega$ resistor. A $3.00\ \mathrm{\mu F}$ capacitor and a $6.00\ \mathrm{\mu F}$ capacitor, joined end to end with each other, are connected across the battery terminals. Everything has been running for a long time.
$3.00\ \mathrm{\mu F}$ in series with $6.00\ \mathrm{\mu F}$, also across the terminals, steady state
Find
(a) Find the current in the resistor and the terminal voltage.
(b) Find the charge on each capacitor.
(c) Find the potential difference across each capacitor.
Hint 1/4
Two separate jobs in one circuit. The capacitor chain carries no current in a steady state, so it does not affect the resistor calculation at all; it simply sits across whatever voltage the terminals settle at.
Hint 2/4
$I = \varepsilon/(R+r)$ and $V_{ab} = \varepsilon-Ir$. For capacitors joined end to end, the same charge sits on every one of them and the reciprocals of the capacitances add; then $V_i = q/C_i$.
Hint 3/4
Here $R+r = 12.0\ \Omega$ with $\varepsilon = 12.0\ \mathrm{V}$, and the capacitor chain is $3.00\ \mathrm{\mu F}$ with $6.00\ \mathrm{\mu F}$ across the terminals.
Hint 4/4
The current is $1.00\ \mathrm{A}$ and the terminals sit at $10.0\ \mathrm{V}$; each capacitor carries $20.0\ \mathrm{\mu C}$, with $6.67\ \mathrm{V}$ and $3.33\ \mathrm{V}$ across them.
Show solutionSolve the resistive circuit, ignoring the capacitors
in the steady state the capacitor chain carries no current, so it contributes nothing to this calculation; using the emf instead of the terminal voltage later would be the standard trap here
with a common charge the smaller capacitance takes the larger voltage, which is the reverse of the resistor case and the reason the two are worth doing side by side
Independent check that the two capacitor voltages add to the terminal voltage: $6.67+3.33 = 10.0\ \mathrm{V}$, which is a constraint the solution never imposed. Had the emf been used instead of the terminal voltage, this sum would have come to $12.0\ \mathrm{V}$ and disagreed with the measured terminals.
Capacitors see the terminal voltage, and in series they share a charge.
2§08.1 — a length of wire as the load●●●○○
Nothing here says which section it belongs to, and that is the point. A copper wire of resistivity $1.68\times10^{-8}\ \Omega\cdot\mathrm{m}$, length $25.0\ \mathrm{m}$ and diameter $1.00\ \mathrm{mm}$ is connected directly across a battery of emf $6.00\ \mathrm{V}$ and internal resistance $0.100\ \Omega$.
(c) Find the power dissipated in the wire and the power wasted inside the battery.
Hint 1/4
The wire has to become a resistance before any circuit reasoning can start. Watch the diameter: the formula wants an area, and the area needs a radius.
Hint 2/4
$R = \rho L/A$ with $A = \pi d^{2}/4$; then $I = \varepsilon/(R+r)$, $V_{ab} = \varepsilon-Ir$ and $P = I^{2}R$ for each element separately.
Hint 3/4
With $\rho = 1.68\times10^{-8}\ \Omega\cdot\mathrm{m}$, $L = 25.0\ \mathrm{m}$ and $d = 1.00\times10^{-3}\ \mathrm{m}$, the area is $\pi(0.500\times10^{-3})^{2}$, and the battery has $\varepsilon = 6.00\ \mathrm{V}$ with $r = 0.100\ \Omega$.
Hint 4/4
The wire is $0.535\ \Omega$, the current is $9.45\ \mathrm{A}$, the terminals sit at $5.05\ \mathrm{V}$, and the powers are $47.8\ \mathrm{W}$ and $8.94\ \mathrm{W}$.
Independent check by the power balance: the battery delivers $\varepsilon I = (6.00)(9.45) = 56.7\ \mathrm{W}$, and $47.8+8.94 = 56.7\ \mathrm{W}$. A second check on the terminal voltage: $IR = (9.45)(0.535) = 5.05\ \mathrm{V}$, which is the same number from the other side.
Halve the diameter before squaring; no thin wire is ideal.
3§08.5 — a parallel plate capacitor sitting in a circuit●●●●○
Again mixed, and again no announcement of which tools it needs. A parallel plate capacitor with plate area $4.00\times10^{-3}\ \mathrm{m^2}$ and an air gap of $0.100\ \mathrm{mm}$ is connected across a resistor of $1.50\ \mathrm{k\Omega}$ that is carrying a steady $2.00\ \mathrm{mA}$ as part of a larger direct-current circuit.
Given
plate area $A = 4.00\times10^{-3}\ \mathrm{m^2}$, gap $d = 1.00\times10^{-4}\ \mathrm{m}$, air between the plates
the capacitor is across a $1.50\ \mathrm{k\Omega}$ resistor
that resistor carries a steady $2.00\ \mathrm{mA}$
Find
(a) Find the capacitance and the potential difference across the capacitor.
(b) Find the charge on the plates and the electric field in the gap.
(c) Find the energy density in the gap, and check it against the total stored energy.
Hint 1/4
The circuit part of this question is one line: the capacitor is across a resistor whose current you know, so its voltage is fixed by Ohm's law. Everything else is geometry.
Hint 2/4
$C = \varepsilon_0A/d$ for a flat pair, $V = IR$ for the resistor, $q = CV$, $E = V/d$ for a uniform gap, and the energy per cubic metre of a field is $u = \tfrac12\varepsilon_0E^{2}$.
Hint 3/4
With $A = 4.00\times10^{-3}\ \mathrm{m^2}$, $d = 1.00\times10^{-4}\ \mathrm{m}$, $\varepsilon_0 = 8.85\times10^{-12}$, and a resistor of $1.50\times10^{3}\ \Omega$ carrying $2.00\times10^{-3}\ \mathrm{A}$.
Hint 4/4
The capacitance is $354\ \mathrm{pF}$ at $3.00\ \mathrm{V}$, holding $1.06\ \mathrm{nC}$ in a field of $3.00\times10^{4}\ \mathrm{V/m}$, with an energy density of $3.98\times10^{-3}\ \mathrm{J/m^3}$.
multiplying by the volume of the gap converts a density into a total, and the gap is the only place the field is
Answer $$\boxed{\,C = 354\ \mathrm{pF},\ V = 3.00\ \mathrm{V},\ q = 1.06\ \mathrm{nC},\ E = 3.00\times10^{4}\ \mathrm{V/m},\ u = 3.98\times10^{-3}\ \mathrm{J/m^3}\,}$$
Check
Independent check on the total energy by the capacitor formula rather than the field: $\tfrac12 CV^{2} = \tfrac12(3.54\times10^{-10})(9.00) = 1.59\times10^{-9}\ \mathrm{J}$, matching the density times the volume. Also, the field is a hundred times below the breakdown field of air, so the gap is safe.
Name the quantity crossing between topics, then solve the halves apart.
4§08.7 — the same ammeter in two different circuits●●●○○
The last of the mixed set. An ammeter of resistance $0.100\ \Omega$ is used twice. First it is inserted into a loop containing a cell of emf $1.50\ \mathrm{V}$ and internal resistance $0.200\ \Omega$ with a $2.00\ \Omega$ resistor. Then the same meter is inserted into a loop with the same cell but a $100\ \Omega$ resistor instead.
external resistor $2.00\ \Omega$ in the first case and $100\ \Omega$ in the second
Find
(a) Find the true current and the reading in the first circuit, and the percentage error.
(b) Find the percentage error in the second circuit.
(c) State in one sentence what property of the circuit decides how badly the meter interferes.
Hint 1/4
Compute the loop resistance twice for each circuit, once without the meter and once with it, and compare the currents. Do not forget that the internal resistance is part of the loop in both.
Hint 2/4
$I = \varepsilon/R_{\rm loop}$, and inserting a meter changes that to $\varepsilon/(R_{\rm loop}+R_A)$, so the ratio of reading to truth is $R_{\rm loop}/(R_{\rm loop}+R_A)$.
Hint 3/4
First circuit: $R_{\rm loop} = 2.00+0.200 = 2.20\ \Omega$ with $R_A = 0.100\ \Omega$. Second: $R_{\rm loop} = 100+0.200 = 100.2\ \Omega$ with the same meter.
Hint 4/4
The first reading is $4.35\%$ low and the second only $0.100\%$ low, because the same meter is a much smaller share of the larger loop.
Independent check on the first case by computing the two currents to more digits and taking the difference directly: $0.681818-0.652174 = 0.029644\ \mathrm{A}$, and $0.029644/0.681818 = 0.0435$, the same $4.35\%$ arrived at through the resistance ratio.
Interference is set by the meter's resistance relative to the circuit's.
Mistake ledger (22 entries)
⚠ Putting only the external resistance in the denominator
the internal resistance is not drawn on most circuit diagrams, and what is not drawn is not added
wrong$$I = \frac{\varepsilon}{R}$$
right$$I = \frac{\varepsilon}{R+r}$$
⚠ Calling the emf the voltage across the load
the number on the battery is the most visible number in the problem, so it gets used for the most visible quantity
wrong$$V_{\rm load} = \varepsilon$$
right$$V_{\rm load} = V_{ab} = \varepsilon - Ir,\qquad \text{equal to } \varepsilon \text{ only when } I = 0$$
⚠ Reporting the power delivered by the source as the power delivered to the load
$\varepsilon I$ is the first product you can form from the given numbers, and it really is a power
wrong$$P_{\rm load} = \varepsilon I$$
right$$P_{\rm load} = I^{2}R = \varepsilon I - I^{2}r$$
⚠ Adding parallel resistors as though they were in series
addition is the rule that gets remembered, and the two pictures are drawn a few centimetres apart in every book
charged from a constant source through any resistance, starting empty
A capacitor at the two ends of the story
$$t = 0:\ \text{empty capacitor behaves as a wire};\qquad t \to \infty:\ \text{behaves as a break}$$
direct-current circuit with constant sources
What a meter does to the circuit
$$\frac{I_{\rm read}}{I_{\rm true}} = \frac{R_{\rm loop}}{R_{\rm loop}+R_A},\qquad R \to \frac{RR_V}{R+R_V}$$
the ammeter is in series with the branch, the voltmeter in parallel with the element
Check yourself
Close the page and write, from memory, the seven things this section can do for you: what a real source hands out and why it is less than the label; the two combination rules and the test that decides which one applies; the order of the reduce and expand procedure; the two rules for a circuit that will not reduce, together with the four signs; the two exponential laws with their time constant; where the energy goes when a capacitor is charged; and what each of the two meters must be. Then write down one circuit for which the combination rules fail, and say what you would do instead.
Given an emf, an internal resistance and a load, produce the current, the terminal voltage and the split of power between load and source, and go backwards from two loaded measurements to the emf and the internal resistance?
c-emf-terminal
Look at any pair of resistors in a drawing and say, with a reason about their ends, whether they are in series, in parallel, or neither, and combine them correctly when they are?
c-series-parallel
Take a network with a real source and produce the current, the voltage and the power in every single element, then verify the whole thing with a power balance?
c-network-reduction
Write down the right number of independent equations for a multi loop circuit, get every sign right on the first attempt, and interpret a negative current without rewriting anything?
c-kirchhoff-rules
Produce the time constant, the final charge and the initial current of a charging circuit, evaluate the charge and current at any stated time, and invert the law to find a time?
c-rc-charging
Convert between a half life and a time constant in both directions, and say how the energy divides between the capacitor and the resistor over a full charge?
c-rc-discharging
Given a meter's resistance, compute how far its reading is from the undisturbed value, and state the resistance it would need in order to be trusted?
c-meters
Glossary (20 terms)
electromotive forceelektromotor kuvvet
The energy a source gives to each coulomb of charge that passes through it, measured in volts and written as a script epsilon. Despite the name it is not a force, and it is equal to the potential difference across the source's terminals only when no current is flowing.
internal resistanceiç direnç
The resistance of the source itself, in series with its emf and inseparable from it. It is what makes the terminal voltage fall as the current rises, and it sets the largest current the source can ever deliver.
terminal voltageuç gerilimi
The potential difference actually available between the two terminals of a source, equal to the emf minus the current times the internal resistance while the source is delivering, and equal to the emf plus that product while the source is being charged.
ideal sourceideal kaynak
A source of zero internal resistance, whose terminal voltage is its emf whatever current is drawn. Real sources approach this only when the current asked of them is small compared with the ratio of emf to internal resistance.
short circuitkısa devre
A connection of negligible resistance placed across a source or an element. Across a source it produces the largest current the source can give, with all the energy dissipated inside the source itself and none of it delivered.
seri bağlantı
An arrangement in which the same current passes through each element in turn, with no junction between them where current could leave. Resistances joined this way add, and the potential differences across them add to the total.
paralel bağlantı
An arrangement in which both ends of every element are joined to the same two conductors, so that all of them carry the same potential difference. The currents through them add, and for resistances the reciprocals add.
equivalent resistanceeşdeğer direnç
The single resistance that would draw the same current from the same source as a whole network. It describes the network only as seen from outside, so the individual currents and voltages inside still have to be recovered one at a time.
branchkol
A path through a circuit between two junctions, carrying one single current along its whole length. The unknowns in a circuit problem are the branch currents, one per branch.
junctiondüğüm
A point at which three or more conductors meet, so that current arriving can divide between several paths. A point where only two wires join is not a junction, since nothing can divide there.
junction ruledüğüm kuralı
The statement that the currents arriving at any junction equal the currents leaving it. It is conservation of charge applied to a point in a steady state, where nothing can accumulate.
loop ruleçevre kuralı
The statement that the potential changes around any closed path add to zero. It holds because the potential at a point is a single number, so a round trip must return to it, and it applies to loops containing no source just as much as to loops containing one.
steady statekararlı durum
The condition a direct-current circuit settles into, in which no current or voltage is still changing and no charge is accumulating anywhere. The only element treated here that has a transient before reaching it is the capacitor.
time constantzaman sabiti
The product of the resistance in the current path and the capacitance, with units of seconds. In one of them a charging capacitor closes sixty three per cent of the gap to its final value, and a discharging one falls to thirty seven per cent of what it had.
RC circuit
A circuit containing a resistance and a capacitance, in which every quantity moves between its initial and final values along an exponential curve rather than jumping. It is the simplest circuit whose behaviour depends on time.
half lifeyarılanma süresi
The time for a decaying quantity to fall to one half of whatever it happened to be, equal to the time constant times the natural logarithm of two. It is the quantity a laboratory measures most easily, and it is shorter than the time constant.
ammeterampermetre
An instrument that measures the current in a branch, and that must therefore be cut into that branch so the same current passes through it. Its own resistance adds to the loop, so it must be as small as possible.
voltmetervoltmetre
An instrument that measures the potential difference between two points, and that is therefore connected across the element rather than into the path. It provides an extra route for current, so its own resistance must be as large as possible.
The condition in which current is driven backwards into the positive terminal of a source by something else in the circuit, so that the source stores energy chemically instead of delivering it. The signature is a terminal voltage above the emf.
open circuitaçık devre
A break in a path, through which no current can pass. An open branch is deleted from the network rather than assigned a very large resistance, and any resistor in series with the break carries no current and therefore drops no voltage.
What comes next
§09 · Magnetism
Everything on this page has been about charge in motion along wires and what pushes it there. The next section asks what a moving charge does to the space around it, which turns out to be something the electric field alone cannot account for.
Sources
Physics for Scientists and Engineers with Modern Physics, D. Giancoli, 5th edition The fixed textbook for this course. The week line for this section names the topic without giving any chapter or section number, so no chapter number is quoted anywhere on this page.
SI units and prefixes All quantities are in ohms, volts, amperes, farads, coulombs, seconds and watts, with prefixes converted to powers of ten before any arithmetic.
Resistivity of copper, 1.68 times ten to the minus eight ohm metres at room temperature Used only in the one interleaved question that hands you a wire rather than a resistance.