Current Electricity Questions and Answers

Answer: B

She can use a steel wire.

A circuit needs a conductor to carry current. Steel is a metal, and metals conduct electricity because they have free electrons, so a steel wire can complete the circuit (though it conducts less well than copper).

Thread, glass and rubber are insulators: they do not let current pass, so the bulb would not glow. Rubber and plastic are used as covering on wires for exactly this reason.

Answer: C

The Assertion is true and the Reason is false.

The filament of an incandescent bulb is made of tungsten. Tungsten has a very high melting point, about 3,400°C, so it can glow white-hot without melting.

The Reason is wrong about resistance. A filament needs a high resistance so that the current produces enough heat to make it glow; tungsten's resistivity is high compared with conductors like copper. Since the Reason states 'low resistance', it is false, which rules out the options calling it true.

Answer: C

The drift velocity is 0.208 mm s⁻¹.

First the current. The wire's resistance is \(R = \frac{L}{\sigma A}\), and with the 1 Ω internal resistance the circuit carries:

\[ \begin{aligned} R &= \frac{100}{(2 \times 10^{8})(0.5 \times 10^{-6})} \\ &= 1\ \Omega \\ I &= \frac{2}{1 + 1} = 1\ \text{A} \end{aligned} \]

Next the electron density, one free electron per atom, with the molar mass 0.0635 kg mol⁻¹:

\[ \begin{aligned} n &= \frac{\rho N_{A}}{M} \\ &= \frac{6.35 \times 10^{3} \times 6 \times 10^{23}}{0.0635} \\ &= 6 \times 10^{28}\ \text{m}^{-3} \end{aligned} \]

Then \(I = neAv_{d}\) gives:

\[ \begin{aligned} neA &= (6 \times 10^{28})(1.6 \times 10^{-19}) \\ &\quad \times (0.5 \times 10^{-6}) = 4800 \\ v_{d} &= \frac{I}{neA} = \frac{1}{4800} \\ &\approx 2.08 \times 10^{-4}\ \text{m s}^{-1} \end{aligned} \]

That is 0.208 mm s⁻¹. The other options are a quarter (0.052), a half (0.104) and three quarters (0.156) of this value. Leaving out either resistance would instead double the current and give about 0.417, which is not offered.

Answer: B

The relation is \(v_{d} \propto E\).

Drift velocity is \(v_{d} = \frac{eE\tau}{m}\), where \(\tau\) is the mean time between collisions. With \(\tau\) independent of the field, \(v_{d}\) is proportional to \(E\).

Then \(j = nev_{d} \propto E\), which is Ohm's law in the form \(j = \sigma E\). Any other dependence (\(E^{2}\), \(\sqrt{E}\), \(\frac{1}{E}\)) would make current non-proportional to voltage, and Ohm's law would fail.

Answer: D

The equivalent resistance is \(\frac{2R}{9}\).

Each side has resistance \(\frac{R}{3}\). Between the ends of one side, that side is in parallel with the other two sides in series (\(\frac{2R}{3}\)):

\[ \begin{aligned} R_{eq} &= \frac{\frac{R}{3} \cdot \frac{2R}{3}}{\frac{R}{3} + \frac{2R}{3}} \\ &= \frac{2R^{2}/9}{R} = \frac{2R}{9} \end{aligned} \]

\(\frac{R}{3}\) is one side alone, \(\frac{2R}{3}\) the other path alone, and \(\frac{R}{9}\) and \(\frac{R}{6}\) come from combining the paths incorrectly.

Answer: B

The net current increases.

For \(n\) identical cells (emf \(E\), internal resistance \(r\)) in parallel, the emf stays \(E\) and the internal resistance drops to \(\frac{r}{n}\). Through an external resistance \(R\):

\[ I = \frac{E}{R + r/n} \gt \frac{E}{R + r} \]

So the current delivered is larger than one cell would give, and each cell supplies only a share of it. The net voltage neither increases nor decreases (it stays \(E\)), and the internal resistance falls rather than rising.

Answer: A

The INCORRECT statement is that Kirchhoff's junction rule is based on conservation of energy.

The junction rule (the sum of currents entering a junction equals the sum leaving) expresses conservation of charge. It is the loop rule (the sum of potential changes around a closed loop is zero) that rests on conservation of energy.

The other statements are correct: Ohm's law gives a linear I-V graph, current is a scalar (it has direction along a wire but does not add as a vector), conductivity is the reciprocal of resistivity, and current density is a vector.

Answer: E

The emf of the cell is 2 V.

The emf is the work done by the cell per unit charge taken round the circuit:

\[ \begin{aligned} \varepsilon &= \frac{W}{q} = \frac{10\ \text{J}}{5\ \text{C}} \\ &= 2\ \text{V} \end{aligned} \]

0.5 V inverts the ratio, and the other values do not follow from the given work and charge.

Answer: C

The resistance of the new wire is \(9R\).

Resistance is \(R = \rho l / A\), proportional to length and inversely proportional to area, for the same material (same \(\rho\)):

\[ \begin{aligned} R' &= \rho \frac{3l}{A/3} = 9 \rho \frac{l}{A} \\ &= 9R \end{aligned} \]

Tripling the length triples the resistance, and cutting the area to a third triples it again. \(3R\) accounts for only one of the two changes, \(R\) assumes they cancel (they reinforce each other), and \(R/3\) and \(R/9\) invert the relationship.

Answer: C

The power dissipated is \(\frac{2V^{2}}{R}\).

Energy dissipated per unit time is power. With a voltage across a resistance, \(P = V^{2}/R\), here with voltage \(2V\) and resistance \(2R\):

\[ \begin{aligned} P &= \frac{(2V)^{2}}{2R} = \frac{4V^{2}}{2R} \\ &= \frac{2V^{2}}{R} \end{aligned} \]

\(\frac{4V^{2}}{R}\) forgets that the resistance is \(2R\). \(\frac{4V}{R}\) does not square the voltage, and \(4VR^{2}\) and \(4V^{2}R^{2}\) have resistance in the numerator, which would make power grow with resistance at fixed voltage, the wrong way round.