NCERT Exemplar Solutions Class 12 Physics Current Electricity — Multiple Choice Questions Q1–11

NCERT Exemplar Problems, Class 12 Physics — Current Electricity, MCQ I and MCQ II. All 11 questions solved.

The book numbers these questions 3.1 to 3.11; here they are Questions 1 to 11. Questions 1 to 6 are MCQ I (exactly one option is correct) and Questions 7 to 11 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Charge conservation. Current is conserved at a junction; momentum is not, and neither is the direction of $\mathbf{j}$.
• Cells. Cells in parallel give an equivalent emf between the individual emfs, weighted by internal resistances.
• Null methods. In the meter bridge and the potentiometer no current flows through the galvanometer at balance, so accuracy depends on the sensitivity of the balance and on the lengths read.
• Conduction. Resistivity is $\rho = m/(ne^2\tau)$, so it depends on carrier number $n$ and relaxation time $\tau$, both of which can change with temperature.

Key insight. Most of these are not calculations but a decision about what is conserved and what is merely a convenient approximation. Questions 7 and 11 turn on it.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

Consider a current carrying wire (current $I$) in the shape of a circle. Note that as the current progresses along the wire, the direction of $\mathbf{j}$ (current density) changes in an exact manner, while the current $I$ remains unaffected. The agent that is essentially responsible for this is

Solution. The source of emf sets up the overall current, but it acts at one place and cannot steer electrons round every bend. What does the steering is a thin distribution of charge on the surface of the wire. It arranges itself so that, at every point, the field inside the wire points along the wire, however it is curved. That field turns $\mathbf{j}$ with the wire while the same current $I$ passes every cross-section. Options (c) and (d) name charges behind and ahead only in the sense of the line of motion, which cannot account for the turning of the direction; it is the surface charge distribution as a whole that does it.

Question 2

Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ $(\varepsilon_2 > \varepsilon_1)$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in Fig. 3.1.

The figure shows two branches between junctions A and B: the upper branch holds the cell $\varepsilon_1$ in series with $r_1$, the lower branch holds $\varepsilon_2$ in series with $r_2$.

Solution. Write the terminal voltage $V$ across AB. With both cells delivering current $I_1$ and $I_2$ out of the branches, $V = \varepsilon_1 – I_1r_1 = \varepsilon_2 – I_2r_2$. Eliminating the currents for the combination (net current $I = I_1 + I_2$) gives

$$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \frac{r_1r_2}{r_1+r_2}.$$

This is a weighted average of $\varepsilon_1$ and $\varepsilon_2$, with positive weights $r_2/(r_1+r_2)$ and $r_1/(r_1+r_2)$ summing to $1$. A weighted average always lies between the two values, so $\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$: (a) is true, (b) is false. The sum in (c) is the result for cells in series, not in parallel. And (d) fails because $\varepsilon_{eq}$ visibly depends on $r_1$ and $r_2$.

Question 3

A resistance $R$ is to be measured using a meter bridge. Student chooses the standard resistance $S$ to be $100\ \Omega$. He finds the null point at $l_1 = 2.9$ cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?

Solution. At balance $R/S = l_1/(100 – l_1)$, so

$$R = 100 \times \frac{2.9}{97.1} \approx 2.99\ \Omega \approx 3\ \Omega.$$

A meter bridge is most accurate when the null point is near the middle of the wire, $l_1 \approx 50$ cm. Near the ends the ratio $l_1/(100-l_1)$ changes violently for a small error in reading $l_1$, and the end corrections matter most. Here the null is at $2.9$ cm, very near the end, because $S$ is about $33$ times larger than $R$. The remedy is to bring $S$ close to $R$: with $S \approx 3\ \Omega$ the null moves to about the middle. Measuring $l_1$ more carefully (a) cannot cure the poor position, and raising $S$ to $1000\ \Omega$ (b) makes it worse by pushing the null even closer to the end.

Question 4

Two cells of emfs approximately $5$ V and $10$ V are to be accurately compared using a potentiometer of length $400$ cm.

Solution. A potentiometer balances an unknown emf against the drop along a length of the wire. For both cells to be balanced, the drop across the whole wire must exceed the larger emf, $10$ V. A $8$ V driver (a) can never balance $10$ V, so no null point exists. In (c), $10$ V across $50$ cm would mean $80$ V across $400$ cm, far more than any driver here. (d) is simply untrue: a potentiometer is the standard instrument for comparing emfs. In (b) the $15$ V driver, with a rheostat $R$ in series reducing the drop across the wire to a little above $10$ V, balances both cells and also keeps the potential gradient small, which improves precision.

Question 5

A metal rod of length $10$ cm and a rectangular cross-section of $1\ \text{cm} \times \tfrac{1}{2}\ \text{cm}$ is connected to a battery across opposite faces. The resistance will be

Solution. $R = \rho L/A$, where $L$ is the distance between the two faces joined to the battery and $A$ is the area of those faces. Compare the three pairings:

The largest $L/A$ comes when current flows the full $10$ cm length through the smallest face, which is (a). The resistance differs by a factor of $400$ between the extremes, so (d) is false.

Question 6

Which of the following characteristics of electrons determines the current in a conductor?

Solution. The current is $I = neAv_d$, which contains the drift velocity $v_d$ and nothing else about the motion. Thermal velocities are of order $10^5$ m/s but are randomly directed, so their average over the electrons is zero and they carry no net charge across a section. Only the small ordered drift, about $10^{-4}$ m/s, produces a net flow. Thermal motion matters indirectly, since it sets the relaxation time $\tau$ and hence how large $v_d$ becomes for a given field, but the current is determined by $v_d$ alone.

Question 7

Kirchhoff’s junction rule is a reflection of

Solution. The junction rule says that the current into a junction equals the current out of it. It holds because charge is conserved, and in steady state a junction does not pile up charge: (b) and (d) are two ways of stating the same thing. Option (a) is false because $\mathbf{j}$ is a vector whose direction changes at a junction, as the current divides among wires pointing different ways. Option (c) is false because charges are scattered and redirected at a junction by the surface charges there; their momentum is not preserved. Only charge, a scalar, is conserved.

Question 8

Consider a simple circuit shown in Fig. 3.2. A symbol of a resistor with an arrow stands for a variable resistance $R’$. $R’$ can vary from $R_0$ to infinity. $r$ is the internal resistance of the battery $(r \ll R \ll R_0)$.

The figure shows a battery $V$ with internal resistance $r$ driving a current $I$ into two parallel branches between A and B: a fixed resistor $R$ and the variable resistance $R’$.

Solution. The parallel combination of $R$ and $R’$ has resistance $R_p = RR’/(R+R’) $, which is always less than $R$ and, because $R’ \ge R_0 \gg R$, is very close to $R$. So the total resistance is almost $r + R$ whatever $R’$ is, and the current $I = V/(r+R_p)$ barely changes. The drop across AB is $IR_p$, also nearly constant: (a) is true, (c) is false. The current through $R’$ is about (drop)$/R’$, which varies directly with $R’$ over a huge range, so (b) is false. Finally, $R_p \le R$ means $r + R_p \le r + R$, so $I = V/(r+R_p) \ge V/(r+R)$, with equality only as $R’ \to \infty$: (d) is true.

Question 9

Temperature dependence of resistivity $\rho(T)$ of semiconductors, insulators and metals is significantly based on the following factors:

Solution. The resistivity is $\rho = m/(ne^2\tau)$. In semiconductors and insulators, raising $T$ frees many more carriers, so $n$ rises sharply and $\rho$ falls: (a). In metals $n$ is nearly fixed, but more vigorous lattice vibrations shorten the time $\tau$ between collisions, so $\rho$ rises with $T$: (b). Thermal expansion of the length (c) is tiny, about $10^{-5}$ per kelvin, and in any case $\rho$ is a property of the material and does not contain the length. The carrier mass (d) does not depend on temperature in this model.

Question 10

The measurement of an unknown resistance $R$ is to be carried out using Wheatstone’s bridge (see Fig. 3.25 of the NCERT book). Two students perform an experiment in two ways. The first student takes $R_2 = 10\ \Omega$ and $R_1 = 5\ \Omega$. The other student takes $R_2 = 1000\ \Omega$ and $R_1 = 500\ \Omega$. In the standard arm, both take $R_3 = 5\ \Omega$. Both find $R = \dfrac{R_2}{R_1}R_3 = 10\ \Omega$ within errors.

Solution. The ratio $R_2/R_1 = 2$ is the same for both students, so both get $R = 10\ \Omega$, but the errors need not match. Since $R = (R_2/R_1)R_3$, any uncertainty in $R_2$, $R_1$ or $R_3$ passes straight into $R$: (b) is true. With arms of $1000\ \Omega$ and $500\ \Omega$ the current in each arm is much smaller for the same battery, so the galvanometer deflection per unit imbalance is tiny, and the exact null point is harder to locate. The first student’s lower resistances give a more sensitive bridge, so the errors are not the same and (a) is false. No instrument is free of error, which rules out (d).

Question 11

In a meter bridge, the point D is a neutral point (Fig. 3.3).

The figure shows the meter wire AC with the resistance $R$ in the left gap and $S$ in the right gap, the common junction B between them joined through the galvanometer G to the jockey D on the wire; the lengths are marked $l_1$ from A to D and $100-l_1$ from D to C. The cell is connected across A and C, with its longer (positive) plate on the A side.

Solution. Treat the potential along the wire. The positive plate of the cell is on the A side, so the potential falls steadily along the wire from A to C. At the neutral point D, the wire is at the same potential as B: $V_D = V_B$.

• (a) The potential along the wire is strictly monotonic, so only one point on it can equal $V_B$. There is no other neutral point. True.
• (b) Left of D, the wire is nearer A, so it is above $V_B$. Current flows from the wire through G to B. True.
• (c) Right of D, the wire is below $V_B$, so current flows from B through G to the wire. True.
• (d) At balance $R/S = l_1/(100 – l_1)$. Increasing $R$ increases the right-hand side, so $l_1$ increases: the neutral point moves to the right, away from A. False.

Printed key error. The key prints (a), (c) and leaves out (b). But (b) and (c) are the same observation made on the two sides of D: the wire is above $V_B$ on one side and below it on the other. If current flows from B to the wire on the right of D, it must flow from the wire to B on the left. They are true or false together, so (a), (c) alone cannot be the answer. The correct answer is (a), (b), (c).

Common mistakes

• Treating the surface charge as a side issue in Question 1. It is the agent that bends the field round the wire, which the emf source alone cannot do.
• Taking $\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$ in Question 2. That is for series; parallel cells give a resistance-weighted average.
• Fixing the reading, not the null position, in Question 3. Accuracy comes from choosing $S$ so the null is near $50$ cm.
• Choosing a driver of $8$ V in Question 4. The drop across the wire must exceed the larger emf, here $10$ V.
• Using the wrong length in Question 5. $L$ is the distance between the connected faces, not the longest side of the rod.
• Adding thermal velocity to the current in Question 6. It is randomly directed and averages to zero.
• Thinking momentum or $\mathbf{j}$ is conserved at a junction in Question 7. Only charge is.
• Leaving out (b) in Question 11. The galvanometer current reverses on the two sides of D; (b) and (c) go together.

Practise next

• Exemplar Current Electricity: Very Short, Short and Long Answer Questions — where the same ideas are applied to junctions, potentiometers, meter bridges and power loss in cables.
• NCERT textbook Chapter 3 — Current Electricity for the theory of drift velocity, cells, Kirchhoff’s rules and the Wheatstone and meter bridges.
• Exemplar Electrostatic Potential and Capacitance — the previous unit, whose potential and conductor ideas carry over.
• Exemplar Moving Charges and Magnetism — the next unit, which puts moving charges and currents in a magnetic field.