Current Electricity — Practice

Question 1

Is electric current a scalar or a vector, and what is the test?

Accepted Answer

A scalar, even though it has an associated direction. The test is that currents meeting at a junction add arithmetically rather than vectorially, so two 3 A currents always give 6 A whatever the angle between the wires. Current density $\vec{J}$, by contrast, is a genuine vector.

Question 2

A wire of resistance 4 $\Omega$ is stretched to three times its length. Find its new resistance.

Accepted Answer

Stretching conserves volume, so tripling the length cuts the area to one third and $R = \rho L/A$ rises by a factor of nine, giving 36 $\Omega$. The resistivity is unchanged — it is a property of the material, not the shape.

Question 3

Why does resistance rise with temperature in a metal but fall in a semiconductor?

Accepted Answer

Both follow from $ho = m/ne^{2}	au$. In a metal the carrier density $n$ is fixed and heating makes the lattice vibrate harder, shortening $	au$ and raising $ho$. In a semiconductor heating promotes many more electrons across the band gap, and that exponential rise in $n$ overwhelms the fall in $	au$, so $ho$ drops.

Question 4

A copper wire of cross-section $10^{-6}$ m$^{2}$ carries 1 A, with $n = 8.5	imes10^{28}$ m$^{-3}$. Find the drift velocity and comment.

Accepted Answer

From $I = neAv_d$, $v_d \approx 7.4	imes10^{-5}$ m s$^{-1}$ — under a tenth of a millimetre per second, so an electron takes about four hours to travel one metre. The lamp still lights instantly because the electric field is established throughout the circuit at nearly light speed, starting every electron drifting at once.

Question 5

A cell of emf 12 V and internal resistance 2 $\Omega$ drives a 4 $\Omega$ resistor. Find the current, terminal voltage and the power wasted inside the cell.

Accepted Answer

$I = 12/6 = 2$ A. Terminal voltage $V = \varepsilon - Ir = 12 - 4 = 8$ V. Internal dissipation is $I^{2}r = 8$ W against $I^{2}R = 16$ W delivered externally, so a third of the total power is lost inside the cell.

Question 6

In a Wheatstone bridge, $P = 10$, $Q = 20$ and $R = 15\ \Omega$. Find $S$ at balance, and state the effect of changing the galvanometer.

Accepted Answer

At balance $P/Q = R/S$, so $S = 30\ \Omega$. Changing the galvanometer has no effect at all: no current flows through it at balance, so its resistance cannot enter the condition. The cell's emf and internal resistance are equally irrelevant.

Question 7

A 60 W and a 100 W bulb, both rated 220 V, are connected in series across 220 V. Which glows brighter and why?

Accepted Answer

Since $R = V^{2}/P$, the 60 W bulb has 807 $\Omega$ and the 100 W bulb 484 $\Omega$. In series the current is common, so $P = I^{2}R$ and the larger resistance dissipates more — the 60 W bulb glows brighter. In parallel the result reverses, because the voltage is then common and $P = V^{2}/R$.

Question 8

Two cells of emf 10 V and 4 V, with internal resistances 1 and 2 $\Omega$, are joined in parallel with like terminals together across a 3 $\Omega$ resistor. Find the external current.

Accepted Answer

Taking $I_1$ and $I_2$ out of the two cells, the loop equations are $4I_1 + 3I_2 = 10$ and $3I_1 + 5I_2 = 4$. Solving gives $I_1 = 3.45$ A and $I_2 = -1.27$ A, so the external current is 2.18 A. The negative $I_2$ means the 4 V cell is being charged by the stronger one — information, not an error.

Question 9

For a 12 V cell of internal resistance 2 $\Omega$, find the external resistance drawing maximum power, the power delivered, and the efficiency. Comment on the result.

Accepted Answer

Maximum transfer occurs at $R = r = 2\ \Omega$, giving $I = 3$ A and an external power of 18 W, with 18 W also dissipated internally. Efficiency is therefore exactly 50 per cent. Maximum power and maximum efficiency are different goals, which is why power distribution keeps source resistance far below load resistance.

Question 10

Explain why a real voltmeter always reads slightly below the true potential difference, and what an ideal one would look like.

Accepted Answer

A voltmeter sits in parallel with the component, so it draws a small current through its own resistance. That current changes the circuit and reduces the very potential difference being measured. An ideal voltmeter would have infinite resistance and draw nothing; the error grows as the voltmeter's resistance approaches that of the component across it.

Question 11

A wire of resistance $R$ is stretched uniformly to three times its original length. Find its new resistance. What if instead it is drawn down to half its original diameter?

Accepted Answer

Stretching conserves volume, so tripling the length divides the area by three and $R' = \rho(3L)/(A/3) = 9R$. Drawing to half the diameter quarters the area, so the length must quadruple, giving $R' = \rho(4L)/(A/4) = 16R$. In general $R \propto L^{2} \propto 1/d^{4}$ at constant volume. The resistivity does not change at all, because the material is the same.

Question 12

An ammeter of resistance 2 Ω is inserted in series with a 10 Ω resistor across a 12 V supply of negligible internal resistance. What does it read, what was the true current, and what error does the insertion introduce?

Accepted Answer

The true current is $12/10 = 1.2$ A. With the ammeter in the loop the total resistance becomes 12 Ω, so it reads $12/12 = 1.0$ A — an error of 17 per cent, and low. An ammeter always reads low because the resistance it adds throttles the very current it is reporting, exactly as a voltmeter reads low by drawing current. The error shrinks as $R_A$ falls toward its ideal of zero.

Practice — Current Electricity

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