A 50 Hz AC current of crest value 1 A flows through the primary coil of a transformer. If the mutual inductance between the primary and secondary coil be 0.5 H, the crest voltage induced in the secondary is?

Given : Frequency of AC f=50Hz and M=0.5H
The crest value of current is Im = 1A
The instantaneous current is given by i = Imsin(ωt) = sin(2πft) = sin(100πt)
The rate of change of current is given by:
didt = ddt(sin(100πt)) = 100πcos(100πt)
The crest value of didt is 100π A/s
The crest voltage in the secondary coil is:
V2 = M(didt)crest = 0.5H × 100π A/s = 50π V ≈ 157 V
The closest option is 150 V. However, there's a discrepancy. Let's re-examine the calculation. The crest value of the current is 1A. The peak-to-peak change in current is 2A. The time period is T = 1/f = 1/50 = 0.02s. The average rate of change of current (over half a cycle) is approximately 2A / (0.02s/2) = 200 A/s. Using this:
V2 = M(didt)avg = 0.5H × 200 A/s = 100V
Considering the instantaneous rate of change, the peak value is ωIm = 2πfIm = 2π(50)(1) = 100π A/s
Therefore:
V2 = M(didt)peak = 0.5H × 100π A/s ≈ 157V
Considering the approximation made and the options given, 150V is a reasonable approximation. The provided solution (100V) seems to have a calculation error using the average rate of change.