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MHT CET202616 April 2026Evening ShiftPhysicsAlternating CurrentActual

Alternating current of peak value ( 2 ) A flows through the primary coil of a transformer. The coefficient of mutual inductance between primary and secondary coils is 1H. The peak value of induced e.m.f. in the secondary coil is (Frequency of a.c. = 50 Hz)

Options

  1. A100 V
  2. B300 V
  3. C200 V
  4. D400 V

Correct answer

C. 200 V

Step-by-step solution

The current in the primary coil is I = I₀ ( t) . The induced e.m.f. in the secondary coil is given by e = -M dI dt . e = -M d dt (I₀ ( t)) = -M I₀ ( t) The peak value of the induced e.m.f. is e₀ = M I₀ . Given M = 1 H, I₀ = 2 A, and f = 50 Hz. The angular frequency is = 2 f = 2 50 = 100 rad/s. Substituting the values, we get: e₀ = 1 ( 2 ) 100 = 200 V. Answer: 200 V

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