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 1 H. The peak em.f. induced in secondary coil is (Frequency of a. c. is 50 Hz)

Step-by-step solution

The alternating current in the primary coil is given by I_p = I₀ ( t) , where I₀ is the peak current and = 2 f . The induced e.m.f. in the secondary coil is given by Faraday's law of induction: e_s = -M dI_p dt e_s = -M d dt (I₀ ( t)) e_s = -M I₀ ( t) The peak e.m.f. induced in the secondary coil is the maximum value of e_s , which is: E₀ = M I₀ Given values are: M = 1 H I₀ = 2 A f = 50 Hz = 2 50 = 100 rad/s Substituting these values into the equation for peak e.m.f.: E₀ = 1 ( 2 ) 100 E₀ = 200 V Answer: 200 V