NEET Alternating Current PYQs & Practice Questions

10 free NEET Physics questions on Alternating Current, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Inductive Reactance · Series LCR Circuit · Ideal Transformer - Turns Ratio and Voltage Ratio · RMS and Peak Values · Phase Relationship in AC Circuits · Series LCR Circuit and Resonance · Power in AC Circuits · Resonant Frequency · Transformers · Quality Factor (Q-factor) · Instantaneous Value of AC Current · Maximum Power Transfer in AC Circuits

1. Q1NEET 2026Series LCR Circuit - Resonance

   An ac voltage V = 220 sin(2x10^3 t) Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L = 10 mH, C = 25 microF, R = 100 Ohm)

   • A. 5.5 A
   • B. 11.0 A
   • C. 22.0 A
   • D. 2.2 A
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   Answer: C — 22.0 A

   Angular frequency omega = 2x10^3 rad/s. XL = omega*L = 2000 x 0.01 = 20 Ohm. XC = 1/(omega*C) = 1/(2000 x 25x10^-6) = 20 Ohm. Since XL = XC, the circuit is at resonance, so impedance Z = R = 100 Ohm. Current amplitude = V0/Z = 220/100 = 2.2 A.

2. Q2NEET 2026Resonance frequency of LCR circuit

   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:

   • A. 15.9 kHz
   • B. 20.7 kHz
   • C. 10.1 kHz
   • D. 13.5 kHz
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   Answer: —

   f=1/(2π√LC)=1/(2π√(1×10⁻³×0.1×10⁻⁶))≈15.9 kHz.

3. Q3NEET 2026Time to reach peak value

   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?

   • A. 1/120 s
   • B. 1/240 s
   • C. 1/30 s
   • D. 1/60 s
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   Answer: A — 1/120 s

   Time to reach peak from zero = T/4 = 1/(4×60) = 1/240 s.

4. Q4NEET 2025Series LCR Circuit

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

   • A. 7.8 A and 30°
   • B. 7.8 A and 45°
   • C. 15.6 A and 30°
   • D. 15.6 A and 45°
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   Answer: A — 7.8 A and 30°

   Z = √(R²+(XL−XC)²) = √(20²+(45−25)²) = √(400+400) = √800. I = V/Z = 220/√800 ≈ 7.8 A. tanθ = (XL−XC)/R = 20/20 = 1, so θ = 45°.

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5. Q5NEET 2024Ideal Transformer - Turns Ratio and Voltage Ratio

   In an ideal transformer, the turns ratio is Np/Ns = 1/2. The ratio Vs:Vp is equal to (the symbols carry their usual meaning):

   • A. 1 : 1
   • B. 1 : 4
   • C. 1 : 2
   • D. 2 : 1
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   Answer: C — 1 : 2

   Np/Ns = Vp/Vs. Given Np/Ns = 1/2, so Vp/Vs = 1/2, giving Vs/Vp = 2/1, i.e., Vs:Vp = 2:1.

6. Q6NEET 2024Peak Current in Capacitive AC Circuit

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

   • A. 1.20 A
   • B. 0.35 A
   • C. 0.58 A
   • D. 0.93 A
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   Answer: C — 0.58 A

   Xc = 1/(ωC) = 1/(2π×50×10×10⁻⁶) ≈ 318.5 Ω. Irms = V/Xc = 210/318.5 ≈ 0.66 A. I0 = Irms×√2 ≈ 0.93 A.

7. Q7NEET 2023LCR Circuit Resonance Frequency

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

   • A. 15.9 kHz
   • B. 1.59 rad/s
   • C. 1.59 kHz
   • D. 15.9 rad/s
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   Answer: B — 1.59 rad/s

   At resonance, ω = 1/√(LC) = 1/√(10×10^-3 × 1×10^-6) = 1/√(10^-8) = 10^4 rad/s. Frequency f = ω/2π = 10^4/(2π) ≈ 1591.5 Hz ≈ 1.59 kHz.

8. Q8NEET 2023Transformer - Primary Current Calculation

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

   • A. 2.7 A
   • B. 3.7 A
   • C. 0.37 A
   • D. 0.27 A
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   Answer: C — 0.37 A

   For an ideal transformer, input power equals output power: 220 × Ip = 60, so Ip = 60/220 = 0.27 A.

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9. Q9NEET 2023Series LCR Circuit Impedance

   The net impedance of circuit (as shown in figure) will be: [Circuit: series LCR with L=50/π mH, C=10^3/π μF, R=10Ω, connected to 220V, 50Hz AC source]

   • A. 15 Ω
   • B. 5√5 Ω
   • C. 25 Ω
   • D. 10√2 Ω
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   Answer: A — 15 Ω

   XL = 2πfL = 2π×50×(50/π)×10^-3 = 5Ω. XC = 1/(2πfC) = 1/(2π×50×(10^3/π)×10^-6) = 10Ω. Z = √(R²+(XL-XC)²) = √(10²+(10-5)²) = √125 = 5√5 Ω.

10. Q10NEET 2022Resonant Frequency of LCR Circuit

    A series LCR circuit with inductance 10H, capacitance 10 μF, resistance 50 Ω is connected to an ac source of voltage, V = 200 sin(100t) volt. If the resonant frequency of the LCR circuit is ν₀ and the frequency of the ac source is ν then:

    • A. ν₀ = ν = 50/π Hz
    • B. ν₀ = 50/π Hz, ν = 50 Hz
    • C. ν = 100 Hz, ν₀ = 100/π Hz
    • D. ν₀ = ν = 5Hz
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    Answer: —

    Resonant frequency ν₀ = 1/(2π√LC) = 1/(2π√(10×10⁻⁵)) = 1/(2π×10⁻²) = 100/2π = 50/π Hz. Comparing V=V₀sinωt with ω=100: ν = ω/2π = 100/2π = 50/π Hz. So ν₀ = ν = 50/π Hz.