Previous year questions: Current Electricity

Ten questions follow, taken from official CBSE papers in the collected corpus. The label gives the paper and the marks. Work each one before reading the answer, and keep the sign convention visible in circuit answers.

Q1 (CBSE Board 2023, 1 mark)

A steady current flows through a metallic wire whose area of cross section increases continuously from one end to the other. How does the magnitude of the drift velocity of the free electrons change along the wire?

Answer: The current is the same at every cross section, because charge does not accumulate in a steady state. Since I = n e A v, and n and e are fixed for the metal, the drift speed must fall as the area grows. So the drift velocity decreases continuously from the thinner end to the thicker end.

Q2 (CBSE Board 2023, 1 mark)

The current density due to drift of electrons in a conductor is given by which expression?

Answer: J = n e v, where n is the number density of free electrons, e is the electronic charge and v is the drift velocity. The direction of J is the direction of conventional current, which is opposite to the drift direction of the electrons. The option n e A v would be the total current, not the current density, because the area A is already divided out.

Q3 (CBSE Board 2023, 1 mark)

The potential difference across a cell in an open circuit is 8 V. It falls to 4 V when a current of 4 A is drawn from it. Find the internal resistance of the cell.

Answer: Open circuit potential difference equals the emf, so E = 8 V. When a current is drawn, the terminal voltage is E - I r. Substitute 4 = 8 - 4 r, which gives 4 r = 4 and r = 1 ohm. The lost 4 V appears as heat inside the cell.

Q4 (CBSE Board 2023, 2 marks)

Two identical cells, each of emf E and internal resistance r, are connected with a load resistance R, first in series and then in parallel. Obtain the condition under which the current through R is the same in both cases.

Answer: In series the current is 2E / (R + 2r). In parallel the equivalent emf is E and the equivalent internal resistance is r/2, so the current is E / (R + r/2). Equate the two: 2/(R + 2r) = 1/(R + r/2). Cross multiplication gives 2R + r = R + 2r, hence R = r. The currents are equal when the load resistance equals the internal resistance of a single cell.

Q5 (CBSE Board 2024, 3 marks)

Define drift velocity. Show that the resistivity of the material of a conductor is inversely proportional to the relaxation time of the free electrons.

Answer: Drift velocity is the average velocity gained by the free electrons of a conductor in the direction opposite to the applied electric field. In the free electron model the drift speed is e E tau / m, so the current density is J = n e^2 E tau / m. Comparing with J = sigma E gives sigma = n e^2 tau / m. Since resistivity is the reciprocal of conductivity, rho = m / (n e^2 tau), which is inversely proportional to the relaxation time. A smaller relaxation time means more frequent collisions and higher resistivity. The same derivation also gives conductivity proportional to the number density, so a material with more free electrons per unit volume conducts better, provided the relaxation time stays comparable.

Q6 (CBSE Board 2024, 3 marks)

What is a Wheatstone bridge? Obtain the necessary conditions under which the Wheatstone bridge is balanced.

Answer: A Wheatstone bridge is a four-arm resistance network with a galvanometer in one diagonal and a cell in the other. At balance the galvanometer shows no deflection, so no current flows through it and its two ends are at the same potential. Let the four resistances be P, Q in the upper arms and R, S in the lower arms. Applying Kirchhoff's loop rule to the two balanced loops gives P/Q = R/S. This is the balance condition, and it lets an unknown resistance be found from three known ones without measuring any current. If the galvanometer is replaced by a cell and the cell by the galvanometer, the balance condition is unchanged, because the bridge is symmetric in its two diagonals. That is the standard reasoning behind interchanging the galvanometer and the battery.

Q7 (CBSE Board 2024, 1 mark)

Four cells, each of emf E and internal resistance r, are connected in series, but one of them is connected with reversed polarity. What are the equivalent emf and equivalent internal resistance of the combination?

Answer: Three cells add their emfs and the reversed cell subtracts, so the equivalent emf is 3E - E = 2E. The internal resistances are in series whatever the polarity, so they add to 4r. The answer is 2E and 4r. Reversing a cell changes the sign of its emf but not its internal resistance. If two cells were reversed instead, the equivalent emf would be zero and no current would flow, which is a useful limiting case to quote in the answer.

Q8 (CBSE Board 2025, 3 marks)

A battery of emf E and internal resistance r is connected to a variable external resistance R. Obtain the expression for the current and the maximum current the battery can supply. Obtain the terminal voltage and its maximum possible value. The battery sends a current I1 when R = R1 and a current I2 when R = R2. Obtain the internal resistance in terms of I1, I2, R1 and R2.

Answer: The current is I = E / (R + r). It is largest when R is smallest, that is when R tends to zero, giving the maximum current E / r. The terminal voltage is V = E - I r = E R / (R + r), and it is largest when the current is smallest, giving V close to E for a very large R. For the two measurements, E = I1(R1 + r) and E = I2(R2 + r). Setting the right sides equal gives I1 R1 + I1 r = I2 R2 + I2 r, so r (I1 - I2) = I2 R2 - I1 R1. Hence r = (I2 R2 - I1 R1) / (I1 - I2). A quick check: if I1 is larger than I2 then R2 must be larger than R1 for a passive cell, and the sign of r comes out positive, which is the physical requirement.

Q9 (CBSE Board 2025, 3 marks)

Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive an expression for the emf and internal resistance of the equivalent cell.

Answer: Let the two cells be connected between points A and B, with the same positive terminal at A. Let the current drawn from the combination be I, with branch currents I1 and I2. The terminal potential difference is the same across both branches: V = E1 - I1 r1 and V = E2 - I2 r2. Also I = I1 + I2. Solve the first two for I1 and I2, substitute into the current equation and compare with V = Eeq - I req. The result is Eeq = (E1 r2 + E2 r1) / (r1 + r2) and req = r1 r2 / (r1 + r2), which reduce to one cell and half the internal resistance when the cells are identical. For two identical cells in parallel the equivalent emf is unchanged and the internal resistance halves, so the current drawn from the combination is larger than from one cell alone.

Q10 (CBSE Additional PQ 2023-24, 2 marks)

Two copper wires P and Q of the same area of cross section are joined in parallel and connected across a battery of potential difference V. If their lengths are in the ratio 1 : 2, find the ratio of the drift velocities of the electrons in the two wires.

Answer: In parallel both wires have the same potential difference V, so the field in each is V / L. The drift speed is e E tau / m, and for the same material the relaxation time is the same, so the drift speed is proportional to the field and hence inversely proportional to the length. Therefore vP / vQ = LQ / LP = 2 / 1. The shorter wire has the larger drift speed. The resistance of the shorter wire is also smaller, so it carries more current, and the two effects are consistent with the same battery voltage across both branches.