Current Electricity Questions and Answers

Answer: E

The mobility is \(\frac{e\tau}{m}\).

Between collisions an electron accelerates at \(\frac{eE}{m}\), so its drift velocity is \(v_{d} = \frac{eE}{m}\tau\). Mobility is drift velocity per unit field:

\[ \mu = \frac{v_{d}}{E} = \frac{e\tau}{m} \]

\(\frac{Ee\tau}{m}\) is the drift velocity itself, not the mobility, and the options with \(\tau^{2}\) have the wrong dependence on the collision time.

Answer: C

The maximum current is 24 A.

The current is greatest when the external resistance is zero (a short circuit), leaving only the internal resistance to limit it:

\[ I_{max} = \frac{E}{r} = \frac{12}{0.5} = 24\ \text{A} \]

12 A uses \(r = 1\ \Omega\), 4 A would need some external resistance, and 16 A and 30 A do not follow.

In practice a battery should never be short-circuited: the whole emf would be dropped across the internal resistance, heating the battery.

Answer: A

The ratio is 2 : 1.

With the battery across AB, the resistor AB (resistance \(R\)) is in parallel with the path ACB, which has two resistors in series (\(2R\)). Both paths have the same 12 V across them, so the currents are inversely proportional to the resistances:

\[ \frac{I_{AB}}{I_{ACB}} = \frac{2R}{R} = 2 \]

1 : 2 inverts the ratio, and 1 : 1 treats the paths as equal.

Answer: E

The value of R is 4 Ω.

The two branches are 2R + 4R = 6R and 4R + 8R = 12R, in parallel.

\[ \begin{aligned} R_{eq} &= \frac{6R \times 12R}{6R + 12R} = 4R \\ R_{eq} &= \frac{V}{I} = \frac{4}{0.25} = 16\ \Omega \\ R &= 4\ \Omega \end{aligned} \]

Treating all four resistors as one series chain (18R = 16) gives about 0.9 Ω, and 8 Ω or 2 Ω come from halving or doubling the parallel combination.

Answer: B

The temperature coefficient is 0.05 per °C.

The coefficient is defined against the resistance at 0 °C: \(R_{t} = R_{0}(1 + \alpha t)\). Write it at both temperatures and divide.

\[ \begin{aligned} \frac{6}{5} &= \frac{1 + 40\alpha}{1 + 30\alpha} \\ 6 + 180\alpha &= 5 + 200\alpha \\ \alpha &= \frac{1}{20} = 0.05\ ^{\circ}\text{C}^{-1} \end{aligned} \]

The value 0.02 comes from the shortcut \(\alpha \approx \frac{R_{2} - R_{1}}{R_{1}(t_{2} - t_{1})} = \frac{1}{50}\), which refers the coefficient to 30 °C instead of 0 °C. With this definition \(R_{0} = 2\ \Omega\), and \(2(1 + 0.05 \times 30) = 5\) checks.

Answer: D

The value is n = 5.

Each piece has resistance \(\frac{25}{n}\). n equal resistors in parallel give one n-th of a single piece.

\[ \begin{aligned} R_{eq} &= \frac{25}{n} \cdot \frac{1}{n} = \frac{25}{n^{2}} = 1 \\ n &= 5 \end{aligned} \]

Forgetting that each piece is shorter (so has less resistance) would give n = 25, not offered; the other values do not satisfy \(n^{2} = 25\).

Answer: A

A Wheatstone bridge is used to measure unknown resistances.

Four resistances P, Q, R and S form a bridge with a galvanometer across it. At balance, when the galvanometer shows no deflection, \(\frac{P}{Q} = \frac{R}{S}\), which gives the unknown resistance from the three known ones. The metre bridge and post office box are practical forms of it.

Currents are measured with ammeters, voltages with voltmeters or potentiometers, and power with wattmeters.

Answer: E

The resistance is \(10^{6}\rho\).

Use \(R = \frac{\rho L}{A}\) in SI units, with \(L = 2000\) m and \(A = 20\ \text{cm}^{2} = 20 \times 10^{-4}\ \text{m}^{2}\):

\[ R = \rho \frac{2000}{2 \times 10^{-3}} = 10^{6}\rho \]

The smaller powers of ten come from leaving the area in cm² or the length in km.

Answer: C

The combination has emf E and internal resistance \(\frac{r}{n}\).

In parallel, all cells share the same terminals, so the emf of the combination is the emf of one cell, E. Their internal resistances are in parallel, giving \(\frac{r}{n}\). The advantage of a parallel grouping is the lower internal resistance, which allows a larger current.

nE and nr describe n cells in series, and the mixed options combine a series rule with a parallel one.

Cells are joined in parallel when the external resistance is small compared with the internal resistance, because the lower combined internal resistance then gives a larger current.

Answer: C

Mobility is the drift velocity per unit electric field.

Mobility is defined as \(\mu = \frac{v_{d}}{E}\), the drift speed acquired by a charge carrier in a field of unit strength. Its SI unit is m² V^-1 s^-1. For electrons in a metal, \(\mu = \frac{e\tau}{m}\), where \(\tau\) is the mean time between collisions.

Drift velocity per unit current, charge, volume or time is not a defined quantity. Current enters only through the current density, j = nev_d, which grows with the drift velocity; it is not what mobility divides by.