Alternating Current NEET PYQs (2005–2026) with Solutions

10 solved Alternating Current PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyLCR Circuits & Resonance

   An AC circuit contains a resistance of 1 kΩ, a capacitor of 0.1 μF and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately:

   1. AOption A: 13.5 kHz
   2. BOption B: 10.1 kHz
   3. COption C: 20.7 kHz
   4. DOption D: 15.9 kHz

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   Correct answer: (D) 15.9 kHz

   Explanation

   For a series LCR circuit, resonance frequency is: Given: L = 1 mH = 10⁻³ H C = 0.1 μF = 10⁻⁷ F Substitute: f₀ = 1 / [2π√(10⁻³ × 10⁻⁷)] = 1 / [2π × 10⁻⁵] ≈ 1.59 × 10⁴ Hz = 15.9 kHz Hence option (D) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasySeries LCR circuit at resonance

   An ac voltage $V = 220\sin(2 \times 10^3 t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given : $L = 10\ \mathrm{mH}$, $C = 25\ \mu\mathrm{F}$, $R = 100\ \Omega$)

   1. AOption A: $2.2\ \mathrm{A}$
   2. BOption B: $5.5\ \mathrm{A}$
   3. COption C: $11.0\ \mathrm{A}$
   4. DOption D: $22.0\ \mathrm{A}$

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   Correct answer: (A) $2.2\ \mathrm{A}$

   Explanation

   $\omega = 2 \times 10^3\ \mathrm{rad\ s^{-1}}$. $X_L = \omega L = 2000 \times 10 \times 10^{-3} = 20\ \Omega$; $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 25 \times 10^{-6}} = 20\ \Omega$. $X_L = X_C$, so the circuit is at resonance and $Z = R = 100\ \Omega$. Current amplitude $I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2\ \mathrm{A}$, option (A).

3. Question 3 (NEET 2026)

   NEET 2026EasyAC Fundamentals

   The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value?

   1. AOption A: 1/120 s
   2. BOption B: 1/60 s
   3. COption C: 1/30 s
   4. DOption D: 1/240 s

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   Correct answer: (D) 1/240 s

   Explanation

   For AC: i=I₀sin(ωt), where ω=2πf=120π rad/s. Peak occurs at sin(ωt)=1 ⇒ ωt=π/2. Therefore t=(π/2)/(120π)=1/240 s. Hence option (D) is correct.

4. Question 4 (NEET 2025)

   NEET 2025MediumLCR Circuits & Resonance

   To an ac power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is respectively -

   1. AOption A: 7.8 A and 30°
   2. BOption B: 7.8 A and 45°
   3. COption C: 15.6 A and 30°
   4. DOption D: 15.6 A and 45°

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   Correct answer: (B) 7.8 A and 45°

   Explanation

   Net reactance X = X_L - X_C = 45 - 25 = 20 Ω. Impedance Z = √(R² + X²) = √(20² + 20²) = 20√2 Ω. Current I = V/Z = 220/(20√2) ≈ 7.8 A. Phase angle tanφ = X/R = 20/20 = 1, so φ = 45°.

5. Question 5 (NEET 2024)

   NEET 2024EasyTransformers

   In an ideal transformer, the turns ratio is N$_p$/N$_s$ = 1/2. The ratio V$_s$ : V$_p$ is equal to (the symbols carry their usual meaning):

   1. AOption A: 1 : 1
   2. BOption B: 1 : 4
   3. COption C: 1 : 2
   4. DOption D: 2 : 1

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   Correct answer: (D) 2 : 1

   Explanation

   For an ideal transformer: $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ Given: $$\frac{N_p}{N_s} = \frac{1}{2}$$ Therefore: $$\frac{N_s}{N_p} = 2$$ Hence: $$V_s : V_p = 2 : 1$$ Therefore, option (D) is correct.

6. Question 6 (NEET 2024)

   NEET 2024MediumAC in Resistor, Inductor & Capacitor

   A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (Take π = 3.14):

   1. AOption A: 1.20 A
   2. BOption B: 0.35 A
   3. COption C: 0.58 A
   4. DOption D: 0.93 A

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   Correct answer: (D) 0.93 A

   Explanation

   Capacitive reactance: $$X_C = \frac{1}{2\pi fC}$$ Given: $$f = 50\,Hz$$ $$C = 10 \times 10^{-6}\,F$$ $$X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}}$$ $$X_C \approx 318\,\Omega$$ RMS current: $$I_{rms} = \frac{V_{rms}}{X_C} = \frac{210}{318} \approx 0.66\,A$$ Peak current: $$I_0 = \sqrt{2}I_{rms}$$ $$I_0 \approx 1.414 \times 0.66 \approx 0.93\,A$$ Hence, option (D) is correct.

7. Question 7 (NEET 2023)

   NEET 2023MediumTransformers

   A 12 V, 60 W lamp is connected to the secondary of a step down transformer, whose primary is connected to mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

   1. AOption A: 0.27 A
   2. BOption B: 2.7 A
   3. COption C: 3.7 A
   4. DOption D: 0.37 A

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   Correct answer: (A) 0.27 A

   Explanation

   For an ideal transformer, input power equals output power. Therefore primary current I = P/V = 60/220 ≈ 0.27 A.

8. Question 8 (NEET 2023)

   NEET 2023MediumAC Fundamentals

   In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 μF and resistance R is 100 Ω. The frequency at which resonance occurs is :

   1. AOption A: 15.9 rad/s
   2. BOption B: 15.9 kHz
   3. COption C: 1.59 rad/s
   4. DOption D: 1.59 kHz

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   Correct answer: (D) 1.59 kHz

   Explanation

   Resonant frequency: f = 1/(2π√LC) Given: L = 10 mH = 10×10⁻³ H C = 1 μF = 10⁻⁶ F f = 1 / (2π√(10×10⁻³ × 10⁻⁶)) = 1 / (2π√10⁻⁸) = 1 / (2π × 10⁻⁴) ≈ 1.59 × 10³ Hz ≈ 15.9 kHz Therefore, option (B) is correct.

9. Question 9 (NEET 2023)

   NEET 2023MediumAC Fundamentals

   The net impedance of circuit (as shown in figure) will be :

   1. AOption A: 10√2 Ω
   2. BOption B: 15 Ω
   3. COption C: 5√5 Ω
   4. DOption D: 25 Ω

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   Correct answer: (C) 5√5 Ω

   Explanation

   Given: $$L=\frac{50}{\pi}\text{ mH}=\frac{50\times10^{-3}}{\pi}\text{ H}$$ $$C=\frac{10^3}{\pi}\,\mu F=\frac{10^{-3}}{\pi}\text{ F}$$ Frequency: $$f=50\text{ Hz}$$ Angular frequency: $$\omega=2\pi f=100\pi$$ Inductive reactance: $$X_L=\omega L=100\pi\times\frac{50\times10^{-3}}{\pi}=5\,\Omega$$ Capacitive reactance: $$X_C=\frac{1}{\omega C}=\frac{1}{100\pi\times\frac{10^{-3}}{\pi}}=10\,\Omega$$ Net reactance: $$X=X_L-X_C=5-10=-5\,\Omega$$ Resistance: $$R=10\,\Omega$$ Impedance: $$Z=\sqrt{R^2+X^2}=\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt{5}\,\Omega$$ Hence, the correct option is (3).

10. Question 10 (NEET 2022)

    NEET 2022EasyAC Fundamentals

    The peak voltage of the AC source is equal to:

    1. AOption A: 1/√2 times the rms value of the AC source
    2. BOption B: The value of voltage supplied to the circuit
    3. COption C: The rms value of the AC source
    4. DOption D: √2 times the rms value of the AC source

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    Correct answer: (D) √2 times the rms value of the AC source

    Explanation

    For alternating voltage: Vᵣₘₛ = V₀/√2 Therefore: V₀ = √2 Vᵣₘₛ Hence option (D) is correct.