Previous year questions: Alternating Current

The questions below come from official CBSE board papers and official Sample Question Papers. The year label in each heading is printed on the source, and the marks are the marks printed for that question or part. Try the question before reading the answer.

Q1 (CBSE Board 2023, 3 marks)

How does resistance differ from impedance? With the help of a suitable phasor diagram, obtain an expression for the impedance of a series LCR circuit connected to a source v = v_m sin omega t.

Answer: Resistance is the opposition offered by a resistor to the current, it is independent of frequency and it dissipates energy as heat. Impedance is the total opposition offered by a circuit containing resistance and reactance together, it depends on frequency and it includes the opposition from energy storage in the inductor and capacitor. For the series LCR circuit the same current flows through all three elements, so take the current phasor as reference. The resistor voltage is in phase with the current, the inductor voltage leads by 90 degrees and the capacitor voltage lags by 90 degrees. The reactive voltages are opposite, so the resultant reactive voltage is proportional to (X_L minus X_C) and perpendicular to the resistive voltage. Adding by the Pythagoras rule gives the voltage amplitude as current times the square root of R squared plus (X_L minus X_C) squared, so Z equals that square root, and the phase angle satisfies tan phi = (X_L minus X_C)/R.

Q2 (CBSE Board 2024, 1 mark)

The reactance of a capacitor of capacitance C connected to an ac source of frequency nu is X. If the capacitance is doubled and the frequency of the source is tripled, the reactance will become (A) X/6 (B) X/3 (C) 3X (D) 6X

Answer: Option (A), X/6. The capacitive reactance is X_C = 1/(2 pi nu C). Doubling the capacitance halves the reactance, and tripling the frequency reduces it to one third, so the combined factor is one sixth. The reactance falls as either the capacitance or the frequency increases.

Q3 (CBSE Board 2024, 4 marks)

(a) A resistor and a capacitor are connected in series to an ac source v = v_m sin omega t. Derive an expression for the impedance of the circuit. (b) When does an inductor act as a conductor in a circuit? Give a reason.

Answer: (a) The same current flows through the resistor and the capacitor, so take the current phasor as reference. The voltage across the resistor is in phase with the current and equals I R in amplitude, while the voltage across the capacitor lags the current by 90 degrees and equals I X_C with X_C = 1/(omega C). The two voltages are perpendicular in the phasor diagram, so the resultant voltage amplitude is I times the square root of R squared plus X_C squared. Dividing by the current gives Z = square root of R squared plus 1/(omega C) squared, and the phase angle has tan phi = X_C/R with the current leading the voltage. (b) An inductor acts as a conductor for direct current, where the frequency is zero, because its reactance X_L = omega L is then zero. It also tends to a conductor at very low frequencies, so only a small voltage appears across it.

Q4 (CBSE Board 2024, 4 marks)

(a) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes of energy loss in a real transformer. (b) A step-up transformer converts a low voltage into a high voltage. Does it violate the principle of conservation of energy? Explain.

Answer: (a) A step-up transformer has a primary of N_p turns and a secondary of N_s greater than N_p turns, both wound on a common laminated soft iron core. The alternating current in the primary sets up an alternating flux in the core, and the secondary picks up that shared flux and develops an emf across its terminals. Each turn sees the same flux, so the emf per turn matches and V_s/V_p = N_s/N_p. The three main losses are copper loss from the resistance of the windings (I squared R), hysteresis loss in magnetising and demagnetising the core each cycle, and eddy current loss in the core, reduced by laminations; flux leakage between the coils is a fourth cause. (b) It does not violate conservation of energy, because an ideal transformer transfers the same power: the secondary voltage is larger, but the secondary current is smaller in the same ratio, so V_s I_s equals V_p I_p. In a real transformer the output power is somewhat less than the input, the difference appearing as heat.

Q5 (CBSE Board 2025, 1 mark)

Assertion (A): A series LCR circuit behaves as a pure resistive circuit at resonance. Reason (R): At resonance, the inductive reactance is equal to the capacitive reactance.

Answer: Both statements are true and the reason correctly explains the assertion, so the answer is option (A). At resonance X_L = X_C, so the two reactances cancel in the impedance formula and Z reduces to R. The voltage and current are then in phase, the power factor is one, and the circuit behaves as if only the resistance were present, which is the pure resistive behaviour stated in the assertion.

Q6 (CBSE Board 2026, 5 marks)

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of the number of turns and the currents in the two coils.

(b) A step-up transformer has a primary to secondary turns ratio of 1:5 and supplies 5 kW of power at 200 V to the primary. Find the primary current and the secondary voltage.

Answer: The diagram shows a laminated soft iron core with the primary coil of N_p turns on one side and the secondary coil of N_s turns on the other, with an alternating source connected to the primary and a load to the secondary. The transformer works on the principle of mutual induction: an alternating current in the primary produces an alternating magnetic flux in the core, and that flux links the secondary and induces an emf there. In an ideal transformer all the flux links both coils, so the emf per turn is the same for each, giving V_s/V_p = N_s/N_p. For a step-up transformer N_s is greater than N_p, so the secondary voltage is larger. With no power loss, V_p I_p = V_s I_s, so I_s/I_p = V_p/V_s = N_p/N_s: the voltage is stepped up by the turns ratio and the current is stepped down by the same factor. (b) With N_s/N_p = 5, the secondary voltage is 5 times the primary voltage, so V_s = 1000 V. The input power is 5 kW, so the primary current is 5000/200 = 25 A, and for an ideal transformer the secondary current is 5000/1000 = 5 A. The current has gone down by the same factor by which the voltage went up, so the power is unchanged and conservation of energy is respected.

Q7 (CBSE SQP 2025-26, 1 mark)

If the phasor diagram for a device connected to an AC supply shows the current phasor leading the voltage phasor by 90 degrees, then which statement is true? (A) When the frequency of the AC source is increased, the impedance of the device decreases. (B) This device behaves as a conducting wire when connected across a DC source. (C) When the frequency of the AC source is decreased, the impedance of the device decreases. (D) The device stores energy in the form of magnetic potential energy.

Answer: Option (A). The current leading the voltage by 90 degrees identifies the device as a capacitor, whose reactance is X_C = 1/(omega C). Increasing the frequency lowers the reactance, so the impedance decreases. A capacitor blocks direct current and stores energy in its electric field, not in a magnetic field, so the other options fail.

Q8 (CBSE SQP 2023-24, 1 mark)

Long distance transmission of electrical energy is carried out with transformers. The generator output is stepped up because of (a) reduction of current (b) reduction of current and voltage both (c) power loss is cut down (d) (a) and (c) both

Answer: Option (d). A step-up transformer raises the voltage and, for the same power, lowers the current in the transmission line. The loss in the line is I squared R, so the smaller current cuts the loss sharply, and the reduced current is also easier to handle in thinner cables. The two effects together are the reason power is transmitted at high voltage.

Q9 (CBSE SQP 2024-25, 1 mark)

In a series LCR circuit, the voltage across the resistance, the capacitance and the inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be (A) 10 V (B) 10 root 2 V (C) 10/root 2 V (D) 20 V

Answer: Option (C), 10/root 2 V, about 7.07 V. Since the voltages across R, C and L are 10 V each, the reactive drops cancel and the supply voltage is 10 V. The equality of the resistor and inductor voltages also gives X_L = R. Shorting the capacitor leaves a series LR circuit with Z = square root of (R squared + X_L squared) = R root 2, so the current is 10/(R root 2). The voltage across the inductor is I X_L = 10/(R root 2) times R = 10/root 2 V. This is what the official marking scheme does.

Q10 (CBSE SQP 2022-23, 3 marks)

An ac source generating a voltage v = v_m sin omega t is connected to a capacitor of capacitance C. Find the expression for the current flowing through it. Plot a graph of v and i versus omega t to show that the current is ahead of the voltage by pi/2.

Answer: The charge on the capacitor is q = C v = C v_m sin omega t, and the current is i = dq/dt = C v_m omega cos omega t. Writing the cosine as a sine shifted by 90 degrees gives i = i_m sin (omega t + pi/2) with i_m = v_m omega C = v_m/X_C. The current therefore leads the voltage by pi/2, which is a quarter of a cycle. The graph shows the current peak occurring a quarter cycle before the voltage peak, and at the moment the voltage is zero the current is maximum. The average power over a full cycle is zero because the current is exactly 90 degrees out of phase with the voltage.