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

According to question, peak value of current
`I_(0)=sqrt(2)xxI_(rms)=(2)/(pi)A`
Coefficient of mutual inductance=1H
As we know, induced emf in secondary coil is given by
`E_(S)=M*(dl)/(dt)" "["where, "I=I_(0)sinomegat]`
`E_(S)=M omega I_(0)cos(omegat)`
`=1xx2pixx50xx(2)/(pi)cos(2pixx50xxt)" "(becauseomegaw=2pin)`
for t=0, we have
`E_(S)=4xx50=200V`