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

Options

A. 100 V
B. 200 V
C. 300 V
D. 400 V

Answer

B

Explanation

Let `I_(p) = I_(0) sin omega t` be the a.c. Applied to the primary.
In this case `I_(0) = 1/(pi)`
`:. I_(P) = 2/(pi) sin (100 pi t) ( :' omega = 2 pi n)`
and `e_(s) = M(dIP)/(dt) = 1 xx d/(dt)[2/(pi)(sin 100 pi t)]`
`=2/(pi) xx 100 pi xx cos (100 pi t)`
`:.` Peak e.m.f. = 200 V
[since maximum value of cos `(100 pi t) = 1]`.