DC Circuits

Drift velocity: the average velocity with which free electrons move in a conductor in the direction opposite to the applied electric field, superimposed on their random thermal motion. It is given by

$v_d=\frac{I}{nAe}$

Wires of different diameter: The same current $I$ flows through both wires, and both are copper, so the free-electron density $n$ is the same. Since $v_d=\dfrac{I}{nAe}$, we get $v_d\propto\dfrac1A\propto\dfrac1{d^2}$.

When the electrons move from the larger-diameter wire into the smaller-diameter wire, the cross-sectional area $A$ decreases, so the drift speed increases. It is inversely proportional to the square of the diameter.

(i) From $P = \dfrac{V^2}{R}$,

$R = \frac{V^2}{P} = \frac{240^2}{480} = 120\ \Omega$

(ii) From $P = VI$,

$I = \frac{P}{V} = \frac{480}{240} = 2\ \text{A}$

Check: $V/R = 240/120 = 2$ A, which agrees.

(b) $4R$

Stretching keeps the volume constant, $AL=\text{const}$. If $L$ doubles, $A$ halves, so $R=\rho L/A$ becomes $\rho(2L)/(A/2)=4R$.

(a) Electric current is the rate of flow of charge through a cross-section of a conductor, $I=\dfrac{\Delta Q}{\Delta t}$. Its SI unit is the ampere (C s$^{-1}$).

(b) From $P=\dfrac{V^2}{R}$, $R=\frac{V^2}{P}=\frac{12^2}{60}=2.4\ \Omega$

Cross-sectional area of the wire: $A=\pi\left(\frac{d}{2}\right)^2=\pi\,(0.40\times10^{-3})^2=5.03\times10^{-7}\ \text{m}^2$

From $R=\dfrac{\rho L}{A}$, $L=\frac{RA}{\rho}=\frac{2.4\times5.03\times10^{-7}}{1.0\times10^{-6}}\approx 1.2\ \text{m}$

(c) Halving the length halves the resistance ($R\propto L$), to $1.2\ \Omega$. Since the supply is still 12 V, $P=V^2/R$ doubles to $120\ \text{W}$. The heater therefore gives twice the power (it also draws twice the current, 10 A instead of 5 A).

One unit of electricity is one kilowatt-hour (kWh), the energy used by a 1 kW appliance running for 1 hour:

$1\ \text{kWh}=1000\ \text{W}\times3600\ \text{s}=3.6\times10^{6}\ \text{J}$

The battery capacity is the charge it can deliver:

$Q=60\ \text{Ah}=60\times3600\ \text{C}=2.16\times10^{5}\ \text{C}$

With negligible internal resistance the terminal voltage stays at $12\ \text{V}$, so

$W=QV=2.16\times10^{5}\times12=2.59\times10^{6}\ \text{J}$

This is $720\ \text{Wh}=0.72\ \text{kWh}$ (equivalently $12\ \text{V}\times5\ \text{A}\times12\ \text{h}$).

Yes. When a cell is being charged, current is driven through it against its emf, so the terminal potential difference is

$V=E+Ir$

which is greater than $E$. When the cell is discharging, $V=E-Ir$, which is less than $E$. On open circuit ($I=0$), $V=E$.

(a) Ohm's law: at constant temperature, the current through a conductor is directly proportional to the potential difference across its ends, $V\propto I$ or $V=IR$.

(b) The maximum resistance comes from connecting all $n$ wires in series: $R_{max}=nr$.

The minimum resistance comes from connecting all of them in parallel: $\dfrac{1}{R_{min}}=\dfrac{n}{r}$, so $R_{min}=\dfrac{r}{n}$.

$\frac{R_{max}}{R_{min}}=\frac{nr}{r/n}=n^2$

The ratio is $n^2:1$.

(a) R/4

The wire is bent at its midpoint, so it forms two equal halves, each of resistance $R/2$. When the two halves are twisted together, they are connected in parallel. Hence the equivalent resistance is

(a) It means the heater converts $2000\ \text{J}$ of electrical energy into heat every second ($2\ \text{kW}=2000\ \text{J s}^{-1}$) at its rated voltage.

(b)

$P=\frac{V^{2}}{R}\ \Rightarrow\ R=\frac{V^{2}}{P}=\frac{220^{2}}{2000}=\frac{48400}{2000}=24.2\ \Omega$

(c) The heat produced per second is $P=\dfrac{V^{2}}{R}$. At the fixed $220\ \text{V}$ supply, power is inversely proportional to resistance, so the resistance of the heating element should be decreased, for example by using a shorter or thicker nichrome wire (since $R=\rho l/A$). A smaller resistance draws more current, which gives more heat per second.

The heater converts 2000 J of electrical energy into heat every second.

From the previous part, the heater is rated $P=2\ \text{kW}=2000\ \text{W}$ and the supply is $V=220\ \text{V}$.

$R=\frac{V^{2}}{P}=\frac{220^{2}}{2000}=\frac{48400}{2000}=24.2\ \Omega$

The resistance of the heater is $24.2\ \Omega$.

Change: decrease the resistance of the heating element, by using a shorter or thicker nichrome wire.

Justification: The heat produced is

$H=\frac{V^{2}}{R}\,t$

At a fixed supply voltage, $H\propto\dfrac{1}{R}$, so a smaller resistance gives more heat in the same time. Since $R=\dfrac{\rho l}{A}$, resistance is lowered by reducing the length $l$ or increasing the cross-sectional area $A$. The current drawn rises, and with it the heating effect $I^{2}Rt$.