Alternating current of peak value 2 π ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

Step-by-step solution

Peak value of current I 0 = I r m s × 2 = 2 π Amp Co-efficient of mutual inductance is M = 1 henry Induced emf in secondary is given by e 2 = M ⋅ d i d t where i = i 0 . sin ⁡ ω t + ϕ Here ω= 2 π n = 100 π . ∴ e 2 = 1 × d d t ( i 0 . sin ⁡ ω t ) i 0 . ω . cos ⁡ ω t = 2 π × 2 π × 50 . cos ⁡ ( 100 π t ) For, t = 0 , we have e 2 = 4 × 50 = 200 V .