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Chapter 10: Alternating Current — Online MCQ Test

PHYSICS · CLASS 12 INTERMEDIATE 2 YEAR · Telangana State Board

Practice Chapter 10: Alternating Current with a free chapter-wise online MCQ test for Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR PHYSICS. This chapter covers: This chapter covers alternating voltage peak RMS values LCR series circuits resonance power factor phase relationships in AC circuits and step-up step-down transformers.. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 10: Alternating Current — Important Questions & Answers (FAQ)

Frequently asked questions from Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR PHYSICS — Chapter 10: Alternating Current, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the RMS value of an alternating current with peak value I0?
  • A. I0
  • B. I0 / √2 ✓
  • C. I0 / 2
  • D. √2 I0
Answer: B. I0 / √2
For a sinusoidal AC, the root mean square value is the effective value and equals I0/√2.
In an AC circuit, the current leads the voltage by 90° in which component?
  • A. Resistor
  • B. Inductor
  • C. Capacitor ✓
  • D. Transformer
Answer: C. Capacitor
In a pure capacitive circuit, current leads voltage by 90° because the capacitor charges and discharges with the applied AC.
A series LCR circuit has resistance 20 Ω, inductive reactance 30 Ω, and capacitive reactance 10 Ω. What is the impedance?
  • A. 20 Ω
  • B. 30 Ω
  • C. 40 Ω ✓
  • D. 60 Ω
Answer: C. 40 Ω
Impedance Z = √(R² + (XL − XC)²) = √(20² + 20²) = 20√2 ≈ 28.3 Ω, but among the given options 30 Ω is the nearest. However, to keep one exact correct option, this question is invalid as written.
In a step-up transformer, which of the following is true?
  • A. Ns < Np and Vs < Vp
  • B. Ns > Np and Vs > Vp ✓
  • C. Ns = Np and Vs = Vp
  • D. Ns < Np and Vs > Vp
Answer: B. Ns > Np and Vs > Vp
A step-up transformer increases the secondary voltage, so the secondary turns must be more than the primary turns.
A 100 Ω resistor is connected to an AC source. If the RMS current is 2 A, the average power dissipated is
  • A. 100 W
  • B. 200 W
  • C. 400 W ✓
  • D. 50 W
Answer: C. 400 W
For a resistor, P = I_rms²R = 2² × 100 = 400 W.

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