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ExamsJEE MainPhysics

In a series LCR a.c. circuit, match the conditions in List-I with the corresponding behaviour in List-II: List-I: (a) omega*L > 1/(omega*C) (b) omega*L = 1/(omega*C) (c) omega*L < 1/(omega*C) (d) At the resonant frequency List-II: (i) Current is in phase with the applied emf (ii) Current lags behind the applied emf (iii) Maximum current flows (iv) Current leads the applied emf

  1. (a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)
  2. (a) - (ii); (b) - (i); (c) - (iii); (d) - (iv)
  3. (a) - (iii); (b) - (i); (c) - (iv); (d) - (ii)
  4. (a) - (iv); (b) - (iii); (c) - (ii); (d) - (i)

Correct answer: (a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)

Solution

When X_L > X_C the circuit is net inductive, so current lags the emf. When X_L = X_C (resonance) the reactances cancel, current is in phase with emf. When X_L < X_C the circuit is net capacitive, so current leads. At resonance impedance Z = R is minimum, so current is maximum. Matching: (a) inductive -> lags (ii); (b) X_L = X_C -> in phase (i); (c) capacitive -> leads (iv); (d) resonance -> maximum current (iii).

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