07Electric current and resistance: from coulombs per second to the bill at the end of the month

Ask what actually has to arrive at the lamp for it to light, and check whether that thing is the same as the electrons that were near the switch.

The current in a metal is $I = nqAv_d$, and the drift speed that comes out of it for an ordinary wire is of order $10^{-4}\ \mathrm{m/s}$, a fraction of a millimetre per second.

Here the wire already contains mobile electrons everywhere along it, including inside the lamp filament, before the switch is touched.

False: the field spreads along the wire almost instantly and sets electrons moving everywhere at once, but each one crawls.

Write down the four quantities that fix the drift speed and cross off the ones that are stated to be identical in the two wires.

From $I = nqAv_d$, the drift speed is $v_d = I/(nqA)$, so with $I$, $n$ and $q$ all fixed, only the area is left.

Here the thin wire has $A_{\rm thin} = A_{\rm thick}/2$, and the area sits in the denominator.

The drift speed is twice as large in the thin wire.

Write down what ohmic means as an equation, then ask what shape of graph that equation forces, and compare it with the shape you were given.

Ohmic means $R = V/I$ is the same number at every point, which forces $I = V/R$, a straight line through the origin.

Here the line is straight but nothing has been said about the origin, and a line such as $I = 0.030V + 0.100$ is perfectly straight while giving a different $V/I$ at every point.

False as stated: straight is not enough, the line also has to pass through the origin.

Before reaching for a formula, decide which of the two quantities in the power expression the question has pinned down.

The voltage is fixed here, so the form whose symbols are both known is $P = V^{2}/R$, and the resistance sits in the denominator.

Here $V$ is unchanged and $R$ becomes $2R$, so the denominator doubles while the numerator stays put.

The power falls to one half.

The quantity asked for in part (a) is on one side of the definition of current and the two quantities you were given are on the other; write the definition and rearrange before substituting anything.

The definition is $I = \Delta Q/\Delta t$, so $\Delta Q = I\,\Delta t$, and a count of carriers is $N = \Delta Q/e$.

Here $I = 4.50\ \mathrm{A}$ and $\Delta t = 6.00\ \mathrm{h} = 2.16\times10^{4}\ \mathrm{s}$, with $e = 1.602\times10^{-19}\ \mathrm{C}$.

The charge is $9.72\times10^{4}\ \mathrm{C}$ and the count is $6.07\times10^{23}$ electrons.

Three parts, and each one uses a different face of the same definition; label them before you start as one instant, one interval average and one interval total.

The instantaneous current is $I = dQ/dt$, the average is $\Delta Q/\Delta t$, and the charge over an interval is the difference of $Q$ at the two ends.

Here $Q(t) = (0.850)t^{3} + (2.40)t$, so $dQ/dt = (2.55)t^{2} + 2.40$, and the values needed are $Q(3.00)$ and $Q(2.00)$.

The answers are $25.4\ \mathrm{A}$, then $10.1\ \mathrm{A}$, then $18.6\ \mathrm{C}$.

The three parts are the same chain each time: turn the thickness into an area, share the current over it, and then ask how fast the carriers have to move to deliver that.

Use $A = \pi d^{2}/4$, then $J = I/A$, then $v_d = J/(ne)$, and finally $t = \ell/v_d$.

Here $d = 1.00\times10^{-3}\ \mathrm{m}$, $I = 3.00\ \mathrm{A}$, $ne = (5.86\times10^{28})(1.602\times10^{-19})\ \mathrm{C/m^{3}}$ and $\ell = 0.100\ \mathrm{m}$.

The answers are $3.82\times10^{6}\ \mathrm{A/m^{2}}$, $4.07\times10^{-4}\ \mathrm{m/s}$ and $246\ \mathrm{s}$.

Two steps and they are in a fixed order: the piece has a resistance before any current is mentioned, so find that first and only then bring the current in.

Use $R = \rho L/A$ with $A = \pi d^{2}/4$, and then the definition of resistance in the form $V = IR$.

Here $L = 45.0\ \mathrm{m}$, $d = 2.60\times10^{-3}\ \mathrm{m}$, $\rho = 2.65\times10^{-8}\ \Omega\cdot\mathrm{m}$ and $I = 8.00\ \mathrm{A}$.

The resistance is $0.225\ \Omega$ and the potential difference is $1.80\ \mathrm{V}$.

Part (b) is not a new calculation; it asks what has to happen to the area when the length is halved and the resistance is to be unchanged.

From $R = \rho L/A$, a fixed $R$ and a fixed $\rho$ mean $L/A$ is fixed, so halving $L$ requires halving $A$, and $A \propto d^{2}$.

Here $R = 40.0\ \Omega$, $\rho = 1.00\times10^{-6}\ \Omega\cdot\mathrm{m}$, $d = 0.250\ \mathrm{mm}$ for part (a), and the halving applies to the area in part (b).

The length is $1.96\ \mathrm{m}$ and the new diameter is $0.177\ \mathrm{mm}$.

The unknown here is inside the bracket rather than outside it, so rearrange for the temperature before putting a single number in.

From $R_T = R_0[1 + \alpha(T - T_0)]$ comes $T = T_0 + (R_T/R_0 - 1)/\alpha$.

Here $R_T/R_0 = 26.5/22.0$, $\alpha = 3.93\times10^{-3}\ (^\circ\mathrm{C})^{-1}$ and $T_0 = 20.0\ ^\circ\mathrm{C}$.

The winding is at $72.0\ ^\circ\mathrm{C}$, a rise of $52.0$ degrees.

Decide which quantity the socket holds fixed before choosing a power formula, and note that the rating already pairs a power with a voltage.

Use $I = P/V$, then $R = V^{2}/P$, and then $E = P\Delta t$ in whichever pair of units the answer is wanted in.

Here $P = 1800\ \mathrm{W}$, $V = 220\ \mathrm{V}$ and $\Delta t = 40.0\ \mathrm{min}$, which is $0.667\ \mathrm{h}$ or $2400\ \mathrm{s}$.

The answers are $8.18\ \mathrm{A}$, $26.9\ \Omega$ and $1.20\ \mathrm{kW\,h}$, that is $4.32\times10^{6}\ \mathrm{J}$.

The question gives a peak and asks mostly for averages, so the very first line has to be the conversion; everything after it is the ordinary power work from earlier in this material.

Use $I_{\rm rms} = I_0/\sqrt{2}$, then $\bar{P} = I_{\rm rms}^{2}R$, then $p_{\rm max} = I_0^{2}R$, and $V_{\rm rms} = I_{\rm rms}R$.

Here $I_0 = 6.00\ \mathrm{A}$, $R = 15.0\ \Omega$ and $\sqrt{2} = 1.414$.

The answers are $4.24\ \mathrm{A}$, $270\ \mathrm{W}$, $540\ \mathrm{W}$ and $63.6\ \mathrm{V}$.

Two things change here and only one of them is stated outright; find the hidden one by asking what constant volume forces on the cross section.

Constant volume gives $A_2L_2 = A_1L_1$, and the resistance obeys $R = \rho L/A$ with $\rho$ unchanged.

Here $L_2 = 2L_1$, so $A_2 = A_1/2$, and both changes go into the ratio $R_2/R_1 = (L_2/L_1)(A_1/A_2)$.

The resistance becomes $4R$.

There are two temperatures in this question and every quantity belongs to exactly one of them; write which temperature each part is asking about before computing anything.

Use $R = V^{2}/P$ at the working temperature, then $R_0 = R_T/[1 + \alpha(T - T_0)]$ to step back, then $L = R_0A/\rho_0$ with $A = \pi d^{2}/4$, and finally $I = V/R_0$ and $P = V^{2}/R_0$ for the cold instant.

Here $V = 220\ \mathrm{V}$, $P = 1200\ \mathrm{W}$, $T - T_0 = 300\ ^\circ\mathrm{C}$, $\alpha = 4.00\times10^{-4}\ (^\circ\mathrm{C})^{-1}$, $d = 0.350\times10^{-3}\ \mathrm{m}$ and $\rho_0 = 1.00\times10^{-6}\ \Omega\cdot\mathrm{m}$.

The answers are $40.3\ \Omega$, $36.0\ \Omega$, $3.46\ \mathrm{m}$, and $6.11\ \mathrm{A}$ with $1.34\ \mathrm{kW}$.

Check the units of each line rather than the arithmetic; one of these steps produces a quantity whose units are not metres per second at all.

The correct relation is $J = nqv_d$, so $v_d = J/(nq)$, and the denominator is a charge per unit volume rather than a number per unit volume.

Here $n = 8.47\times10^{28}\ \mathrm{m^{-3}}$ and $q = e = 1.602\times10^{-19}\ \mathrm{C}$, so $nq = 1.357\times10^{10}\ \mathrm{C/m^{3}}$ and $J = 4.53\times10^{6}\ \mathrm{A/m^{2}}$.

Step 3 is the first wrong one; the drift speed is $3.34\times10^{-4}\ \mathrm{m/s}$ and the time is about $3.00\times10^{3}\ \mathrm{s}$.

Notice that the lead and the kettle do not have the same quantity held fixed, so they do not take the same power formula; decide separately for each.

For the kettle at its rating use $I = P/V$; for the lead use $R = \rho L/A$ and then $P = I^{2}R$, because what the lead has in common with the kettle is the current and not the voltage.

Here $P = 2200\ \mathrm{W}$, $V = 220\ \mathrm{V}$, $L = 40.0\ \mathrm{m}$, $d = 1.29\times10^{-3}\ \mathrm{m}$ and $\rho = 1.68\times10^{-8}\ \Omega\cdot\mathrm{m}$.

The answers are $10.0\ \mathrm{A}$, $0.514\ \Omega$, $51.4\ \mathrm{W}$ which is $2.34$ per cent, and $5.14\ \mathrm{V}$ lost.

Two different chapters meet here: one of them tells you how many carriers arrive per second, the other tells you what each one brings with it.

The count per second is $N = I/e$, the energy each electron gains crossing a potential difference is $W = eV$, and a rate of energy delivery is a power.

Here $I = 12.0\times10^{-3}\ \mathrm{A}$, $V = 25.0\times10^{3}\ \mathrm{V}$ and $e = 1.602\times10^{-19}\ \mathrm{C}$.

The answers are $7.49\times10^{16}$ per second, $4.01\times10^{-15}\ \mathrm{J}$ each, and $300\ \mathrm{W}$.

Part (b) asks for a field inside a conductor, which sounds like it contradicts an earlier result; recall what condition that earlier result carried and check whether this wire meets it.

Use $J = I/A$, then the microscopic form $E = \rho J$, and then the uniform field relation $V = EL$ from the potential material.

Here $A = \pi(2.00\times10^{-3})^{2}/4$, $I = 5.00\ \mathrm{A}$, $\rho = 1.68\times10^{-8}\ \Omega\cdot\mathrm{m}$ and $L = 100\ \mathrm{m}$.

The answers are $1.59\times10^{6}\ \mathrm{A/m^{2}}$, $0.0267\ \mathrm{V/m}$ and $2.67\ \mathrm{V}$.

The first part belongs to the capacitor material and the second to this one; identify which of the given numbers each part actually uses before writing anything.

Use $Q = CV$ for the stored charge, $I = \Delta Q/\Delta t$ for the current, $U = \tfrac{1}{2}CV^{2}$ for the stored energy and $\bar{P} = U/\Delta t$ for the average power.

Here $C = 25.0\times10^{-6}\ \mathrm{F}$, $V = 200\ \mathrm{V}$ and $\Delta t = 5.00\times10^{-3}\ \mathrm{s}$.

The answers are $5.00\times10^{-3}\ \mathrm{C}$, $1.00\ \mathrm{A}$, $0.500\ \mathrm{J}$ and $100\ \mathrm{W}$.

Three chapters meet in this one question, and each part belongs to exactly one of them; label the parts before starting.

Use $V = Ed$ for a uniform field, $C = \varepsilon_0 A/d$ and $Q = CV$ for the plates, and $I = \Delta Q/\Delta t$ for the delivery.

Here $E = 3.00\times10^{6}\ \mathrm{V/m}$, $d = 1.00\times10^{-3}\ \mathrm{m}$, $A = 0.0200\ \mathrm{m^{2}}$ and $\Delta t = 20.0\times10^{-3}\ \mathrm{s}$.

The answers are $3.00\times10^{3}\ \mathrm{V}$, $5.31\times10^{-7}\ \mathrm{C}$ and $2.66\times10^{-5}\ \mathrm{A}$.