The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 25 — Current, Resistance & Electromotive Force

Problems with worked solutions · 15 points · 3 extra credit

Chapter 25 — Current, Resistance & Electromotive Force

Core: 15 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (2 pts ••) A copper wire of diameter $2.05\,\text{mm}$ carries a steady current of $8.0\,\text{A}$. Copper has $n=8.5\times10^{28}$ free electrons per cubic metre.

   (a) Find the current density $J$ in the wire.

   (b) Find the drift speed of the conduction electrons.

   (c) How many electrons pass through a cross-section of the wire in one minute?

   Solution

   1.

   (a) Cross-sectional area: $A=\pi (d/2)^2=\pi(1.025\times10^{-3})^2\approx 3.30\times10^{-6}\,\text{m}^2$.

   $$J=\frac{I}{A}=\frac{8.0}{3.30\times10^{-6}}\approx 2.4\times10^{6}\,\text{A/m}^2.$$

   (b) Drift speed $v_d=\dfrac{J}{n e}=\dfrac{2.42\times10^{6}}{(8.5\times10^{28})(1.60\times10^{-19})}\approx 1.8\times10^{-4}\,\text{m/s}$ (about $0.18\,\text{mm/s}$ — very slow).

   (c) Charge in one minute: $q=I\,t=(8.0)(60)=480\,\text{C}$. Number of electrons $=q/e=480/(1.60\times10^{-19})\approx 3.0\times10^{21}$ electrons.

2. 2. (2 pts ••) A heating element is made from nichrome wire ($\rho=1.10\times10^{-6}\,\Omega\!\cdot\!\text{m}$) of length $3.5\,\text{m}$ and diameter $0.50\,\text{mm}$.

   (a) Find the resistance of the element at room temperature.

   (b) A second element uses the same material but with twice the length and half the diameter. By what factor does its resistance change relative to part (a)? Give the new resistance.

   Solution

   2.

   (a) $A=\pi(0.25\times10^{-3})^2\approx 1.96\times10^{-7}\,\text{m}^2$.

   $$R=\frac{\rho L}{A}=\frac{(1.10\times10^{-6})(3.5)}{1.96\times10^{-7}}\approx 20\,\Omega.$$

   (b) $R=\rho L/A$ with $A\propto d^2$, so $R\propto L/d^2$. Doubling $L$ multiplies $R$ by $2$; halving $d$ divides $A$ by $4$, multiplying $R$ by $4$. Net factor $=2\times4=8$. New resistance $\approx 8\times19.6\approx 1.6\times10^{2}\,\Omega$ ($\approx157\,\Omega$).

3. 3. (2 pts ••) A copper coil has a resistance of $25.0\,\Omega$ at $20\,{}^\circ\text{C}$. The temperature coefficient of resistivity for copper is $\alpha=3.9\times10^{-3}\,\text{K}^{-1}$, and over this range you may treat resistance as varying linearly with temperature.

   (a) Find the resistance of the coil when it heats up to $85\,{}^\circ\text{C}$.

   (b) At what temperature would the coil's resistance read $30.0\,\Omega$?

   Solution

   3.

   (a) $R=R_0[1+\alpha(T-T_0)]=25.0\,[1+(3.9\times10^{-3})(85-20)]=25.0(1+0.2535)\approx 31\,\Omega$.

   (b) Solve $30.0=25.0[1+\alpha(T-20)]$: $\;\dfrac{30.0}{25.0}-1=0.20=\alpha(T-20)$, so $T-20=0.20/3.9\times10^{-3}\approx 51\,\text{K}$, giving $T\approx 71\,{}^\circ\text{C}$.

4. 4. (1 pt •) A resistor is connected across an ideal $12\,\text{V}$ source and obeys Ohm's law with $R=470\,\Omega$.

   (a) Find the current through the resistor.

   (b) Find the power it dissipates.

   Solution

   4.

   (a) $I=\dfrac{V}{R}=\dfrac{12}{470}\approx 2.6\times10^{-2}\,\text{A}$ ($\approx26\,\text{mA}$).

   (b) $P=\dfrac{V^2}{R}=\dfrac{12^2}{470}\approx 0.31\,\text{W}$.

5. 5. (2 pts ••) A single AA cell has an emf of $1.50\,\text{V}$ and internal resistance $r=0.35\,\Omega$. It is connected to an external resistor $R=4.5\,\Omega$.

   (a) Find the current in the circuit.

   (b) Find the terminal voltage of the cell.

   (c) Find the power dissipated in the internal resistance and the power delivered to $R$, and confirm they sum to the total power the emf supplies.

   Solution

   5.

   (a) Current: $I=\dfrac{\mathcal{E}}{R+r}=\dfrac{1.50}{4.5+0.35}=\dfrac{1.50}{4.85}\approx 0.31\,\text{A}$.

   (b) Terminal voltage: $V_{ab}=\mathcal{E}-Ir=1.50-(0.309)(0.35)\approx 1.39\,\text{V}$. (Equivalently $IR=0.309\times4.5\approx1.39\,\text{V}$.)

   (c) Power in internal resistance: $P_r=I^2 r=(0.309)^2(0.35)\approx 0.033\,\text{W}$. Power to the load: $P_R=I^2 R=(0.309)^2(4.5)\approx 0.43\,\text{W}$. Sum $\approx 0.464\,\text{W}$, which equals the emf's output $\mathcal{E}I=(1.50)(0.309)\approx 0.46\,\text{W}$. ✓

6. 6. (2 pts ••) A household space heater is rated $1500\,\text{W}$ when operated at $120\,\text{V}$.

   (a) Find the operating resistance of the heater and the current it draws.

   (b) If the heater runs $2.5$ hours per day for $30$ days and electricity costs \0.14$ per kilowatt-hour, find the total energy used (in kWh) and the cost.

   Solution

   6.

   (a) $R=\dfrac{V^2}{P}=\dfrac{120^2}{1500}=9.6\,\Omega$; current $I=\dfrac{P}{V}=\dfrac{1500}{120}=12.5\,\text{A}$.

   (b) Energy $=P\times t=1.5\,\text{kW}\times(2.5\,\text{h/day}\times30\,\text{day})=1.5\times75=112.5\,\text{kWh}$. Cost =112.5\times\0.14\approx $16$.

7. 7. (2 pts ••) In a cylindrical conductor of radius $R=1.2\,\text{mm}$, the current is not uniform: the current density points along the axis and has magnitude $J(r)=J_0\,(r/R)$, where $r$ is the distance from the axis and $J_0=1.5\times10^{6}\,\text{A/m}^2$.

   (a) By integrating over the cross-section, find the total current carried by the conductor. (Hint: use a thin annular ring of radius $r$ and thickness $dr$.)

   (b) Find the average current density $I/(\pi R^2)$ and compare it to $J_0$.

   Solution

   7.

   (a) Take a thin annular ring of radius $r$, thickness $dr$, area $dA=2\pi r\,dr$. The current through it is $dI=J(r)\,dA=J_0\dfrac{r}{R}\,(2\pi r\,dr)$. Integrate from $0$ to $R$:

   $$I=\int_0^R \frac{2\pi J_0}{R}\,r^2\,dr=\frac{2\pi J_0}{R}\cdot\frac{R^3}{3}=\frac{2\pi J_0 R^2}{3}.$$

   Numerically: $I=\dfrac{2\pi(1.5\times10^{6})(1.2\times10^{-3})^2}{3}\approx 4.5\,\text{A}$.

   (b) Average current density $=\dfrac{I}{\pi R^2}=\dfrac{2\pi J_0 R^2/3}{\pi R^2}=\dfrac{2}{3}J_0=1.0\times10^{6}\,\text{A/m}^2$. It is $\tfrac{2}{3}$ of $J_0$, because the density is smallest near the centre (where there is little area) and largest at the rim.

8. 8. (2 pts ••) A battery of emf $\mathcal{E}$ and internal resistance $r=0.50\,\Omega$ drives an external resistor $R$. When $R=3.0\,\Omega$ the current is $2.4\,\text{A}$.

   (a) Find the emf $\mathcal{E}$.

   (b) The external resistor is then removed and the battery's terminals are short-circuited (external resistance $\to 0$). Find the short-circuit current.

   Solution

   8.

   (a) $\mathcal{E}=I(R+r)=2.4\,(3.0+0.50)=2.4\times3.5=8.4\,\text{V}$.

   (b) Short circuit means external resistance $\to0$, so $I_{sc}=\dfrac{\mathcal{E}}{r}=\dfrac{8.4}{0.50}\approx 17\,\text{A}$.

9. Extra Credit (3 pts •••)

   A solid conductor is shaped as a truncated cone (frustum) of length $L$. Its circular cross-section has radius $a$ at the left face and grows linearly to radius $b>a$ at the right face. The material has uniform resistivity $\rho$, and current flows along the axis. Assume the current density is uniform across each cross-section (a good approximation when the taper is gentle), so each thin disc acts as a resistor in series.

   (a) Let $x$ measure distance from the left face. Write the radius $\varrho(x)$ of the cross-section as a function of $x$, and hence the resistance $dR$ of a thin disc of thickness $dx$.

   (b) Integrate to show that the total resistance is

   $$R=\frac{\rho L}{\pi a b}.$$

   (c) Evaluate $R$ for a copper frustum ($\rho=1.7\times10^{-8}\,\Omega\!\cdot\!\text{m}$) with $L=0.15\,\text{m}$, $a=1.0\,\text{mm}$, and $b=3.0\,\text{mm}$. Check that your formula reduces to $\rho L/(\pi a^2)$ when $b\to a$ (a uniform cylinder).

   Solution

   Extra Credit

   (a) The radius grows linearly from $a$ at $x=0$ to $b$ at $x=L$:

   $$\varrho(x)=a+\frac{b-a}{L}\,x.$$

   A thin disc of thickness $dx$ at position $x$ has cross-sectional area $\pi\varrho(x)^2$ and, treating it as a small resistor,

   $$dR=\frac{\rho\,dx}{\pi\,\varrho(x)^2}=\frac{\rho\,dx}{\pi\left(a+\frac{b-a}{L}x\right)^2}.$$

   (b) Discs in series add, so $R=\displaystyle\int_0^L dR$. Substitute $u=a+\frac{b-a}{L}x$, so $du=\frac{b-a}{L}\,dx$, i.e. $dx=\frac{L}{b-a}\,du$; limits $u:a\to b$:

   $$R=\frac{\rho}{\pi}\cdot\frac{L}{b-a}\int_a^b \frac{du}{u^2} =\frac{\rho L}{\pi(b-a)}\left[-\frac{1}{u}\right]_a^b =\frac{\rho L}{\pi(b-a)}\left(\frac{1}{a}-\frac{1}{b}\right).$$

   Since $\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{b-a}{ab}$, the $(b-a)$ factors cancel:

   $$R=\frac{\rho L}{\pi(b-a)}\cdot\frac{b-a}{ab}=\boxed{\dfrac{\rho L}{\pi a b}}.$$

   (c) $R=\dfrac{(1.7\times10^{-8})(0.15)}{\pi(1.0\times10^{-3})(3.0\times10^{-3})}=\dfrac{2.55\times10^{-9}}{9.42\times10^{-6}}\approx 2.7\times10^{-4}\,\Omega$ ($\approx0.27\,\text{m}\Omega$).

   Check: as $b\to a$, $R\to\dfrac{\rho L}{\pi a^2}$, the familiar uniform-cylinder result — the geometric mean $\sqrt{ab}$ of the two radii plays the role of the effective radius (since $ab=(\sqrt{ab})^2$). ✓