Chapter-7 Alternating Current — Online MCQ Test
PHYSICS · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter-7 Alternating Current with a free chapter-wise online MCQ test.
This chapter covers: AC circuits - RMS value - reactance - impedance - LCR circuit - resonance - transformer - AC generator.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-7 Alternating Current — Important Questions & Answers
What is the RMS (root mean square) value of an alternating current with peak value I₀?
- A. I₀/2
- B. I₀/√2
- C. I₀√2
- D. 2I₀
Answer: B. I₀/√2
The RMS value of an AC current is defined as I₀/√2, where I₀ is the peak (maximum) value of the current.
The RMS value of an AC current is defined as I₀/√2, where I₀ is the peak (maximum) value of the current.
Which of the following is the SI unit of reactance?
- A. Ampere
- B. Ohm
- C. Henry
- D. Farad
Answer: B. Ohm
Reactance (inductive or capacitive) is measured in Ohms (Ω), the same unit as resistance.
Reactance (inductive or capacitive) is measured in Ohms (Ω), the same unit as resistance.
A sinusoidal AC voltage is given by V(t) = 230sin(ωt) V. What is its RMS voltage?
- A. 230 V
- B. 230/√2 V ≈ 163 V
- C. 230√2 V ≈ 325 V
- D. 115 V
Answer: B. 230/√2 V ≈ 163 V
The peak voltage is 230 V, so RMS voltage = 230/√2 ≈ 163 V. The value 230 V used in India is already the RMS value.
The peak voltage is 230 V, so RMS voltage = 230/√2 ≈ 163 V. The value 230 V used in India is already the RMS value.
In a series RLC circuit, if the inductance is decreased while keeping resistance and capacitance constant, the resonant frequency will:
- A. Decrease
- B. Increase
- C. Remain unchanged
- D. Become zero
Answer: B. Increase
Resonant frequency f₀ = 1/(2π√LC) is inversely proportional to √L. Decreasing L increases the resonant frequency.
Resonant frequency f₀ = 1/(2π√LC) is inversely proportional to √L. Decreasing L increases the resonant frequency.
A resonant LCR circuit has R = 5 Ω, L = 10 mH, and the resonant frequency is 1000 Hz. What is the capacitance?
- A. 2.53 μF
- B. 25.3 μF
- C. 0.253 μF
- D. 253 μF
Answer: A. 2.53 μF
At resonance, f₀ = 1/(2π√LC), so C = 1/(4π²f₀²L) = 1/(4π² × 10⁶ × 0.01) ≈ 2.53 × 10⁻⁶ F = 2.53 μF.
At resonance, f₀ = 1/(2π√LC), so C = 1/(4π²f₀²L) = 1/(4π² × 10⁶ × 0.01) ≈ 2.53 × 10⁻⁶ F = 2.53 μF.