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Alternating Current: JEE Main 2026 PYQs

Every 2026 Session 2 question on this chapter, in one place.

Last updated: 26 October 2026

Direct answer

JEE Main 2026 asked 5 questions on Alternating Current across 5 shifts (Session 2).

Revise the chapter first: Alternating Current notes and formulas. Then solve these without looking at solutions.

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Alternating Current questions

Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.

2 April 2026 Shift 2, Q42 · Q42 · MCQ · Alternating Current · Medium

The figure given below shows an $LCR$ series circuit with two switches $S_1$ and $S_2$. When switch $S_1$ is closed keeping $S_2$ open, the phase difference ($\phi$) between the current and source voltage is $30°$ and phase difference is $60°$ when $S_2$ is closed keeping $S_1$ open. The value of $(3L_1 - L_2)$ is ______ H. [Figure: An LCR circuit with $C = 100\mu F$, $R$, two inductors $L_1$ and $L_2$ with switches $S_1$ and $S_2$, and voltage source $v = V_0 \sin(300t)$]

  1. (1) $\frac{9}{2}$
  2. (2) $\frac{2}{9}$
  3. (3) $\frac{1}{3}$
  4. (4)
Show answer and solution

Answer: Option (2)

Key idea: Phase difference in LCR circuits and impedance analysis

  1. Calculate capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{300 \times 100 \times 10^{-6}} = \frac{100}{3}$ Ω
  2. For Case 1 (S₁ closed): $\tan(30°) = \frac{X_{L1} - X_C}{R}$ gives $\frac{1}{\sqrt{3}} = \frac{300L_1 - \frac{100}{3}}{R}$
  3. For Case 2 (S₂ closed): $\tan(60°) = \frac{X_{L2} - X_C}{R}$ gives $\sqrt{3} = \frac{300L_2 - \frac{100}{3}}{R}$
  4. Divide the two equations: $\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = 3 = \frac{300L_2 - \frac{100}{3}}{300L_1 - \frac{100}{3}}$
  5. Cross-multiply and simplify: $900L_1 - 100 = 300L_2 - \frac{100}{3}$
  6. Rearrange: $900L_1 - 300L_2 = 100 - \frac{100}{3} = \frac{200}{3}$
  7. Solve for $(3L_1 - L_2)$: $300(3L_1 - L_2) = \frac{200}{3}$, so $3L_1 - L_2 = \frac{2}{9}$ H

4 April 2026 Shift 1, Q50 · Q50 · Numerical · Alternating Current · Medium

An inductor of 10 mH, capacitor of 0.1 $\mu$F and a resistor of 100 $\Omega$ are connected in series across an $a.c$ power supply 220 V, 70 Hz. The power factor of the given circuit is 0.5. The difference in the inductive reactance and capacitance reactance is $\sqrt{3}$ u $\Omega$. The value of u is ______.

Show answer and solution

Answer: 100

Key idea: Power factor in RLC series circuit and impedance calculation

  1. Given: $L = 10$ mH, $C = 0.1$ μF, $R = 100$ Ω, power factor $= 0.5$
  2. Power factor $\cos\phi = \frac{R}{Z}$, so $0.5 = \frac{100}{Z}$ gives $Z = 200$ Ω
  3. For series RLC circuit: $Z^2 = R^2 + (X_L - X_C)^2$
  4. Substitute: $200^2 = 100^2 + (X_L - X_C)^2$ gives $(X_L - X_C)^2 = 30000$
  5. Therefore $|X_L - X_C| = \sqrt{30000} = 100\sqrt{3}$ Ω
  6. Given $|X_L - X_C| = \sqrt{3} \times u$, so $100\sqrt{3} = \sqrt{3} \times u$
  7. Solving gives $u = 100$

5 April 2026 Shift 1, Q44 · Q44 · MCQ · Alternating Current · Medium

An a.c. source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current is observed as $I$. Now the frequency of the source is changed to $\omega/4$, (keeping the voltage unchanged) the current is found to be $I/3$. The ratio of resistance to reactance at frequency $\omega$ is

  1. (1) $\sqrt{\frac{6}{7}}$
  2. (2) $\sqrt{\frac{3}{5}}$
  3. (3) $\sqrt{\frac{7}{8}}$
  4. (4)
Show answer and solution

Answer: Option (3)

Key idea: RC Circuit Impedance and Frequency Dependence

  1. For RC series circuit: $Z = \sqrt{R^2 + X_C^2}$ where $X_C = \frac{1}{\omega C}$
  2. At frequency $\omega$: $V = I\sqrt{R^2 + \frac{1}{\omega^2 C^2}}$
  3. At frequency $\omega/4$: Reactance becomes $X_C' = 4X_C$, so $V = \frac{I}{3}\sqrt{R^2 + \frac{16}{\omega^2 C^2}}$
  4. Since voltage is constant: $\sqrt{R^2 + \frac{1}{\omega^2 C^2}} = \frac{1}{3}\sqrt{R^2 + \frac{16}{\omega^2 C^2}}$
  5. Squaring and simplifying: $9R^2 + \frac{9}{\omega^2 C^2} = R^2 + \frac{16}{\omega^2 C^2}$
  6. This gives: $8R^2 = \frac{7}{\omega^2 C^2}$, so $\frac{R^2}{X_C^2} = \frac{7}{8}$
  7. Therefore: $\frac{R}{X_C} = \sqrt{\frac{7}{8}}$

5 April 2026 Shift 2, Q48 · Q48 · Numerical · Alternating Current · Easy

A series LCR circuit with $R = 20$ $\Omega$, $L = 1.6$ H and $C = 40$ $\mu$F is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is __________ $\Omega$.

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Answer: 200

Key idea: Resonance in LCR circuits - At resonance, inductive reactance equals capacitive reactance

  1. Given: $R = 20\,\Omega$, $L = 1.6\,\text{H}$, $C = 40\,\mu\text{F} = 40 \times 10^{-6}\,\text{F}$
  2. At resonance, the angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$
  3. Calculate: $\omega_0 = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125\,\text{rad/s}$
  4. Inductive reactance: $X_L = \omega_0 L = 125 \times 1.6 = 200\,\Omega$
  5. Verify: At resonance $X_L = X_C$, so $X_C = \frac{1}{\omega_0 C} = \frac{1}{125 \times 40 \times 10^{-6}} = 200\,\Omega$ ✓

6 April 2026 Shift 1, Q45 · Q45 · MCQ · Alternating Current · Easy

A LCR series circuit driven with $E_{rms} = 90$ V at frequency $f_d = 30$ Hz has resistance $R = 80$ $\Omega$, an inductance with inductive reactance $X_L = 20.0$ $\Omega$ and capacitance with capacitive reactance $X_C = 80.0$ $\Omega$. The power factor of the circuit is_____.

  1. (1) 0.8
  2. (2) 0.54
  3. (3) 0.9
  4. (4) 0.5
Show answer and solution

Answer: Option (1)

Key idea: Power factor in LCR series circuit

  1. Identify the given values: $R = 80\,\Omega$, $X_L = 20\,\Omega$, $X_C = 80\,\Omega$
  2. Calculate the net reactance: $X_L - X_C = 20 - 80 = -60\,\Omega$
  3. Calculate the impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + 60^2}$
  4. Simplify: $Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$
  5. Calculate power factor: $\cos\phi = \frac{R}{Z} = \frac{80}{100} = 0.8$
FAQs

Frequently asked questions

How many questions come from Alternating Current in JEE Main?

In the 9 shifts of JEE Main 2026 Session 2 we found 5 questions on it, about 0.6 per shift.

Where can I practise Alternating Current PYQs for free?

On this page, with solutions. For more, use NTA Abhyas and our free mock tests.

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