"Current Electricity" — Practice

Question 1

Six identical resistors of resistance $R$ form the edges of a tetrahedron. Find the resistance between any two vertices.

Accepted Answer

Label the terminals $A$ and $B$, and the other two vertices $C$ and $D$. Symmetry makes $V_C=V_D$, so the resistor between them carries no current and can be removed.

What remains: the direct $AB$ resistor of $R$, in parallel with two paths $A$-$C$-$B$ and $A$-$D$-$B$, each of $2R$.

$\frac{1}{R_{eq}}=\frac1R+\frac{1}{2R}+\frac{1}{2R}=\frac2R$

$R_{eq}=\dfrac R2$

Question 2

An infinite ladder has series resistors of $R$ and shunt resistors of $2R$. Find the input resistance.

Accepted Answer

$X=R+\dfrac{2RX}{2R+X}$

$X\left(2R+Xight)=R\left(2R+Xight)+2RX$

$X^{2}-RX-2R^{2}=0$

$X=\dfrac{R\pm3R}{2}$, and taking the positive root, $X=2R$.

The rare exact answer here is a coincidence of these particular values; most ladders give an irrational multiple of $R$.

Question 3

Twelve resistors of $6\ \Omega$ each form a cube. Find the resistance across the body diagonal, and the current in each edge when $9$ V is applied across it.

Accepted Answer

$R_{eq}=\dfrac{5R}{6}=\dfrac{5	imes6}{6}=5\ \Omega$

Total current $=\dfrac{9}{5}=1.8$ A.

The three edges at each end each carry $\dfrac{1.8}{3}=0.6$ A.

The six middle edges each carry $\dfrac{1.8}{6}=0.3$ A.

Checking the voltage: $0.6	imes6+0.3	imes6+0.6	imes6=3.6+1.8+3.6=9$ V.

Question 4

A conductor of resistivity $\rho$ is shaped as a truncated cone of end radii $a$ and $3a$ and length $L$. Find its resistance, and the percentage error in using the mean radius.

Accepted Answer

$R=\dfrac{ho L}{\pi\left(aight)\left(3aight)}=\dfrac{ho L}{3\pi a^{2}}$

The mean radius is $2a$, giving $\dfrac{ho L}{4\pi a^{2}}$.

Ratio $=\dfrac{1/3}{1/4}=1.333$, so the mean-radius estimate is **low by $25\%$**.

The geometric mean of the end areas is what matters, and it is always smaller than the area at the mean radius.

Question 5

A hollow sphere of inner radius $a$ and outer radius $b$ is made of material of resistivity $\rho$. Current flows radially outward. Find the resistance.

Accepted Answer

$dR=\dfrac{ho\,dr}{4\pi r^{2}}$

$R=\dfrac{ho}{4\pi}\displaystyle\int_a^{b}\frac{dr}{r^{2}}=\dfrac{ho}{4\pi}\left(\frac1a-\frac1bight)$

As $b	o\infty$ this tends to the finite value $\dfrac{ho}{4\pi a}$, so a sphere embedded in an infinite conducting medium has finite resistance to infinity — the basis of earthing electrode design.

Question 6

A battery of emf $24$ V and internal resistance $4\ \Omega$ supplies a variable load $R$. Find the load for maximum power, the maximum power, and the load at which the power delivered is half its maximum.

Accepted Answer

Maximum at $R=r=4\ \Omega$, giving $P_{max}=\dfrac{576}{16}=36$ W.

For half power: $\dfrac{576R}{\left(R+4ight)^{2}}=18$

$32R=\left(R+4ight)^{2}\qquad\Rightarrow\qquad R^{2}-24R+16=0$

$R=12\pm\sqrt{128}=23.3\ \Omega$ or $0.686\ \Omega$ — two loads, one above and one below the matched value.

Question 7

A $20\ \mu$F capacitor is charged through a $5\ 	ext{k}\Omega$ resistor from a $100$ V supply. Find the time constant, the time to reach $63\%$ of full charge, and the total heat dissipated.

Accepted Answer

$	au=RC=\left(5	imes10^{3}ight)\left(2	imes10^{-5}ight)=0.1$ s

The charge reaches $1-e^{-1}=63\%$ of its final value after exactly one time constant, so $t=0.1$ s.

$W_{dissipated}=	frac12CV^{2}=	frac12\left(2	imes10^{-5}ight)\left(10^{4}ight)=0.1$ J

The same $0.1$ J is stored on the capacitor, and the battery supplied $0.2$ J in total.

Question 8

A battery of emf $\varepsilon$ and negligible internal resistance is connected through $R_1$ to a parallel combination of $R_2$ and an uncharged capacitor $C$. Find the current through the battery immediately after closing the switch and long afterwards.

Accepted Answer

**At $t=0$:** the uncharged capacitor is a short circuit, so $R_2$ is bypassed entirely.

$i_0=\dfrac{\varepsilon}{R_1}$

**At $t	o\infty$:** the capacitor is fully charged and passes no current.

$i_{\infty}=\dfrac{\varepsilon}{R_1+R_2}$

The current falls with time constant $\dfrac{R_1R_2C}{R_1+R_2}$, which uses the parallel combination as seen by the capacitor.

Question 9

A lamp is marked $60$ W, $240$ V. Find its working resistance and, taking $\alpha=4	imes10^{-3}$ K$^{-1}$ with a temperature rise of $1800$ K, its resistance and current at the moment of switching on.

Accepted Answer

$R_{hot}=\dfrac{V^{2}}{P}=\dfrac{57600}{60}=960\ \Omega$, running current $0.25$ A.

$R_0=\dfrac{960}{1+\left(4	imes10^{-3}ight)(1800)}=\dfrac{960}{8.2}=117\ \Omega$

Switch-on current $=\dfrac{240}{117}=2.05$ A

about eight times the running current, which is why the filament so often fails at switch-on.

Question 10

A galvanometer of resistance $50\ \Omega$ shows full-scale deflection for $1$ mA. Find the shunt needed for a $2$ A ammeter, and the effective resistance of the resulting ammeter.

Accepted Answer

$S=\dfrac{I_gG}{I-I_g}=\dfrac{\left(10^{-3}ight)(50)}{2-0.001}=\dfrac{0.05}{1.999}=0.025\ \Omega$

The ammeter's resistance is $G$ in parallel with $S$:

$R_A=\dfrac{50	imes0.025}{50.025}=0.025\ \Omega$

Essentially the shunt alone, which is exactly what a good ammeter needs so that inserting it barely changes the circuit.

Question 11

A cell of emf $1.5$ V and internal resistance $0.5\ \Omega$ is measured by a voltmeter of resistance $49.5\ \Omega$, and by a potentiometer. Find both readings and explain the difference.

Accepted Answer

Voltmeter: $I=\dfrac{1.5}{50}=0.03$ A, so it reads $0.03	imes49.5=1.485$ V.

Potentiometer: at balance no current flows from the cell, so it reads the full $1.50$ V.

The difference of $0.015$ V is the drop across the internal resistance. The voltmeter is not wrong: it correctly reports the terminal voltage while the cell is delivering current.

Question 12

A wire of area $2$ mm$^{2}$ carries $10$ A and is joined to a wire of the same material of area $0.5$ mm$^{2}$. Compare the current, current density, drift speed and electric field in the two sections.

Accepted Answer

Current is **identical**, $10$ A, since charge cannot accumulate at the junction.

$J=I/A$ is four times larger in the thin wire: $5	imes10^{6}$ versus $2	imes10^{7}$ A m$^{-2}$.

$v_d=J/ne$ is likewise **four times** larger there.

$E=J/\sigma$ is also four times larger, which is why the thin section dissipates four times the power per unit volume and runs correspondingly hotter.

Practice — "Current Electricity"

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