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📘 Alternating Current — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Alternating Current for JEE Main with previous year questions on AC circuits, resonance, and transformers. Each solution is explained in a simple, step-by-step manner to boost your score.
Alternating Current — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 2)
An inductor of reactance \(100 \, \Omega\), a capacitor of reactance \(50 \, \Omega\), and a resistor of resistance \(50 \, \Omega\) are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is ________ W.
✅ Answer & Explanation:
Correct Answer: 1
- Total impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (100 - 50)^2} = 50\sqrt{2} \, \Omega\).
- RMS current \(I_{rms} = V_{rms} / Z = 10 / (50\sqrt{2}) = \frac{1}{5\sqrt{2}} \, \text{A}\).
- Average power \(P = I_{rms}^2 R = \left(\frac{1}{5\sqrt{2}}\right)^2 \times 50 = \frac{1}{50} \times 50 = 1 \, \text{W}\).
Q2. JEE Main 2025 (4 April Shift 2)
An inductor of self inductance 1 H connected in series with a resistor of \(100\pi \, \Omega\) and an ac supply of \(100\pi\) volt, 50 Hz. Maximum current flowing in the circuit is ________ A.
✅ Answer & Explanation:
Correct Answer: 1
- Angular frequency \(\omega = 2\pi f = 2\pi(50) = 100\pi \, \text{rad/s}\).
- Inductive reactance \(X_L = \omega L = 100\pi \times 1 = 100\pi \, \Omega\).
- Impedance \(Z = \sqrt{R^2 + X_L^2} = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi\sqrt{2} \, \Omega\).
- Peak voltage \(V_0 = V_{rms}\sqrt{2} = 100\pi\sqrt{2} \, \text{V}\).
- Peak current \(I_0 = V_0 / Z = (100\pi\sqrt{2}) / (100\pi\sqrt{2}) = 1 \, \text{A}\).
Q3. JEE Main 2025 (23 Jan Shift 2)
In a series LCR circuit, a resistor of \(300 \, \Omega\), a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is _____ \(\times 10^4 \, \text{radians s}^{-1}\).
✅ Answer & Explanation:
Correct Answer: 2
- Maximum current occurs at resonance: \(\omega_r = \frac{1}{\sqrt{LC}}\).
- \(\omega_r = \frac{1}{\sqrt{0.1 \times 25 \times 10^{-9}}} = \frac{1}{\sqrt{2.5 \times 10^{-9}}} = \frac{1}{\sqrt{25 \times 10^{-10}}}\).
- \(\omega_r = \frac{1}{5 \times 10^{-5}} = 0.2 \times 10^5 = 2 \times 10^4 \, \text{rad/s}\).
- The value is 2.
Q4. JEE Main 2025 (22 Jan Shift 2)
A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is \(I_0\). If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be :
(1) \(2I_0\) (2) \(I_0\)
(3) \(I_0/2\) (4) \(I_0/\sqrt{2}\)
(1) \(2I_0\) (2) \(I_0\)
(3) \(I_0/2\) (4) \(I_0/\sqrt{2}\)
✅ Answer & Explanation:
Correct Option: (3)
- At resonance, impedance \(Z = R\). Current amplitude \(I_0 = E_0 / R\).
- If \(R' = 2R\), then \(I' = E_0 / 2R = I_0 / 2\).
Q5. JEE Main 2023 (30 Jan Shift 1)
In a series LR circuit with \(X_L = R\), power factor is \(P_1\). If a capacitor of capacitance C with \(X_C = X_L\) is added to the circuit the power factor becomes \(P_2\). The ratio of \(P_1\) to \(P_2\) will be :
(1) 1 : 3 (2) 1 : \(\sqrt{2}\)
(3) 1 : 1 (4) 1 : 2
(1) 1 : 3 (2) 1 : \(\sqrt{2}\)
(3) 1 : 1 (4) 1 : 2
✅ Answer & Explanation:
Correct Option: (2)
- Case 1: \(P_1 = R / \sqrt{R^2 + X_L^2} = R / \sqrt{R^2 + R^2} = 1/\sqrt{2}\).
- Case 2: Adding C with \(X_C = X_L\) makes the circuit resonant (\(X_{net} = 0\)). \(P_2 = R / R = 1\).
- Ratio \(P_1 / P_2 = (1/\sqrt{2}) : 1 = 1 : \sqrt{2}\).
Q6. JEE Main 2022 (28 Jun Shift 1)
A telegraph line of length 100 km has a capacity of \(0.01 \, \mu\text{F/km}\) and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____ mH. (If \(\pi = \sqrt{10}\))
✅ Answer & Explanation:
Correct Answer: 100
- Total capacitance \(C = 100 \times 0.01 = 1 \, \mu\text{F}\).
- Frequency \(f = 500 \, \text{Hz} \implies \omega = 1000\pi\).
- Min impedance \(\implies\) resonance \(\implies L = 1/(\omega^2 C)\).
- \(L = \frac{1}{(1000\pi)^2 \times 10^{-6}} = \frac{1}{10^6 \times \pi^2 \times 10^{-6}} = 1/\pi^2\).
- Since \(\pi^2 = 10\), \(L = 0.1 \, \text{H} = 100 \, \text{mH}\).
Q7. JEE Main 2022 (27 Jul Shift 2)
A series LCR circuit has \(L = 0.01 \, \text{H}, R = 10 \, \Omega\) and \(C = 1 \, \mu\text{F}\) and it is connected to ac voltage of amplitude 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately :
(1) 466 mA (2) 312 mA
(3) 238 mA (4) 196 mA
(1) 466 mA (2) 312 mA
(3) 238 mA (4) 196 mA
✅ Answer & Explanation:
Correct Option: (3)
- Resonant angular frequency \(\omega_0 = 1/\sqrt{LC} = 10^4 \, \text{rad/s}\).
- Frequency 60% lower means \(\omega = 0.4 \omega_0 = 4000 \, \text{rad/s}\).
- Calculating \(Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}\) at this frequency leads to a current amplitude of approximately 238 mA.
Q8. JEE Main 2021 (26 Aug Shift 1)
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance \(R = 3 \, \text{k}\Omega\), an inductor of inductive reactance \(X_L = 250\pi \, \Omega\) and an unknown capacitor. The value of capacitance to maximise the average power should be _______ \(\mu\text{F}\). (take \(\pi^2 = 10\))
✅ Answer & Explanation:
Correct Answer: 4
- Power is maximum at resonance: \(X_C = X_L = 250\pi\).
- \(1 / (2\pi f C) = 250\pi \implies 1 / (100\pi C) = 250\pi\).
- \(C = 1 / (25000\pi^2) = 1 / (25000 \times 10) = \frac{1}{250000} = 4 \times 10^{-6} \, \text{F} = 4 \, \mu\text{F}\).
Q9. JEE Main 2025 (3 April Shift 2)
An electric bulb rated as 100 W — 220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :
(1) 0.64 A (2) 0.45 A
(3) 2.2 A (4) 0.32 A
(1) 0.64 A (2) 0.45 A
(3) 2.2 A (4) 0.32 A
✅ Answer & Explanation:
Correct Option: (1)
- RMS current \(I_{rms} = P / V_{rms} = 100 / 220 = \frac{5}{11} \, \text{A} \approx 0.454 \, \text{A}\).
- Peak current \(I_0 = I_{rms} \times \sqrt{2} = 0.454 \times 1.414 \approx 0.64 \, \text{A}\).
Q10. JEE Main 2025 (29 Jan Shift 1)
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \((R/\sqrt{R^2 + \omega^2 L^2})\), where \(\omega\) is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true but (R) is not the correct explanation of (A)
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \((R/\sqrt{R^2 + \omega^2 L^2})\), where \(\omega\) is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true but (R) is not the correct explanation of (A)
✅ Answer & Explanation:
Correct Option: (2)
The choke coil limits the current in the tube circuit by providing large inductive reactance with minimal resistive power loss. The voltage across the resistive part (tube) is indeed reduced by the power factor \(\cos \phi = \frac{R}{Z}\), which protects the device.
The choke coil limits the current in the tube circuit by providing large inductive reactance with minimal resistive power loss. The voltage across the resistive part (tube) is indeed reduced by the power factor \(\cos \phi = \frac{R}{Z}\), which protects the device.
Q11. JEE Main 2024 (27 Jan Shift 2)
The primary side of a transformer is connected to 230 V, 50 Hz supply. The turn ratio of primary to secondary winding is 10 : 1. Load resistance connected to the secondary side is \(46 \, \Omega\). The power consumed in it is:
(1) 12.5 W (2) 10.0 W
(3) 11.5 W (4) 12.0 W
(1) 12.5 W (2) 10.0 W
(3) 11.5 W (4) 12.0 W
✅ Answer & Explanation:
Correct Option: (3)
- Secondary voltage \(V_s = V_p \times (N_s / N_p) = 230 \times (1 / 10) = 23 \, \text{V}\).
- Power consumed \(P = V_s^2 / R_s = 23^2 / 46 = 529 / 46 = 11.5 \, \text{W}\).
Q12. JEE Main 2024 (06 Apr Shift 1)
When a dc voltage of 100 V is applied to an inductor, a dc current of 5 A flows through it. When an ac voltage of 200 V peak value is connected to inductor, its inductive reactance is found to be \(20\sqrt{3} \, \Omega\). The power dissipated in the circuit is _______ W.
✅ Answer & Explanation:
Correct Answer: 250
- Resistance of coil \(R = V_{dc} / I_{dc} = 100 / 5 = 20 \, \Omega\).
- Impedance \(Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = 40 \, \Omega\).
- \(V_{rms} = V_0 / \sqrt{2} = 200 / \sqrt{2}\).
- Power \(P = V_{rms}^2 R / Z^2 = \frac{(200/\sqrt{2})^2 \times 20}{40^2} = \frac{20000 \times 20}{1600} = 250 \, \text{W}\).
Q13. JEE Main 2023 (01 Feb Shift 1)
A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contain a resistance \(R = 100 \, \Omega\) and an inductor of inductive reactance \(X_L = 79.6 \, \Omega\). The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be ______ \(\mu\text{F}\).
✅ Answer & Explanation:
Correct Answer: 40
- Energy rate is maximized at resonance: \(X_C = X_L = 79.6 \, \Omega\).
- \(1 / (2\pi f C) = 79.6 \implies C = \frac{1}{2\pi \times 50 \times 79.6} = \frac{1}{100\pi \times 79.6}\).
- \(C \approx \frac{1}{314.16 \times 79.6} \approx 40 \times 10^{-6} \, \text{F} = 40 \, \mu\text{F}\).
Q14. JEE Main 2022 (25 Jun Shift 1)
Match List-I with List-II:
Choose the correct answer: (1) (A)-II, (B)-I, (C)-IV, (D)-III
| List - I | List - II |
|---|---|
| (A) AC generator | (I) Detects presence of current |
| (B) Galvanometer | (II) Mech energy to Elec energy |
| (C) Transformer | (III) Principle of resonance |
| (D) Metal detector | (IV) Changes alternating voltage |
Choose the correct answer: (1) (A)-II, (B)-I, (C)-IV, (D)-III
✅ Answer & Explanation:
Correct Option: (1)
- Generator: Conversion of energy [II].
- Galvanometer: Current detection [I].
- Transformer: Voltage regulation [IV].
- Metal detector: Uses LC resonance shifts [III].
Q15. JEE Main 2024 (29 Jan Shift 2)
In the given figure, the charge stored in \(6 \, \mu\text{F}\) capacitor, when points A and B are joined by a connecting wire is _______ \(\mu\text{C}\).
✅ Answer & Explanation:
Correct Answer: 36
When A and B are joined, the potentials at these points equalize. Analyzing the steady-state potential distribution across the capacitor branch relative to the supply voltage (9V) leads to a stored charge of 36 \(\mu\text{C}\).
When A and B are joined, the potentials at these points equalize. Analyzing the steady-state potential distribution across the capacitor branch relative to the supply voltage (9V) leads to a stored charge of 36 \(\mu\text{C}\).
Q16. JEE Main 2021 (22 Jul Shift 1)
In a circuit consisting of a capacitance and a generator with alternating emf, which phasor diagram is correct?
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (3)
In a purely capacitive circuit, the current leads the voltage by \(90^\circ\) (\(\pi/2\) radians). Phasor (3) correctly represents this leading relationship.
In a purely capacitive circuit, the current leads the voltage by \(90^\circ\) (\(\pi/2\) radians). Phasor (3) correctly represents this leading relationship.
Q17. JEE Main 2020 (07 Jan Shift 1)
A LCR circuit behaves like a clamped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:
(1) \(L \leftrightarrow m, C \leftrightarrow k, R \leftrightarrow b\) (2) \(L \leftrightarrow 1/b, C \leftrightarrow 1/m, R \leftrightarrow 1/k\)
(3) \(L \leftrightarrow k, C \leftrightarrow b, R \leftrightarrow m\) (4) \(L \leftrightarrow m, C \leftrightarrow 1/k, R \leftrightarrow b\)
(1) \(L \leftrightarrow m, C \leftrightarrow k, R \leftrightarrow b\) (2) \(L \leftrightarrow 1/b, C \leftrightarrow 1/m, R \leftrightarrow 1/k\)
(3) \(L \leftrightarrow k, C \leftrightarrow b, R \leftrightarrow m\) (4) \(L \leftrightarrow m, C \leftrightarrow 1/k, R \leftrightarrow b\)
✅ Answer & Explanation:
Correct Option: (4)
- Comparing the differential equations:
- Mechanical: \(m \ddot{x} + b \dot{x} + kx = 0\)
- Electrical: \(L \ddot{q} + R \dot{q} + (\frac{1}{C})q = 0\)
- Equivalences: \(L \leftrightarrow m\), \(R \leftrightarrow b\), and \(\frac{1}{C} \leftrightarrow k \implies C \leftrightarrow \frac{1}{k}\).
📘 Exam Preparation Tip for Alternating Current
- RMS Values: For a sinusoidal AC, \(I_{rms} = I_0/\sqrt{2}\) and \(V_{rms} = V_0/\sqrt{2}\). Power in AC circuits is calculated using RMS values.
- Impedance: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\) where \(X_L = \omega L\) and \(X_C = 1/\omega C\). The phase angle \(\phi = \tan^{-1}((X_L - X_C)/R)\).
- Power in AC Circuits: \(P = V_{rms} I_{rms} \cos\phi = I_{rms}^2 R\). Only the resistor dissipates power; reactive components (L and C) store and return energy.
- Resonance: Occurs when \(X_L = X_C\) (\(\omega_r = 1/\sqrt{LC}\)). At resonance, impedance is minimum (\(Z = R\)), current is maximum, and power factor is unity (\(\cos\phi = 1\)).
- Transformer: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}\) (ideal transformer). Power is conserved: \(V_p I_p = V_s I_s\).
- LC Oscillations: The energy oscillates between the capacitor (electric field) and the inductor (magnetic field). Angular frequency \(\omega = 1/\sqrt{LC}\).
- Choke Coil: Used to reduce current in AC circuits without dissipating significant power (since it has low resistance).