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

Answer

B

Explanation

According to question, peak value of current
`l_0 =sqrt2xxl_(rms)=2/piA`
Coefficient of mutual inductance 1H As we know, Induced emf in secondary coil is given by
`E_s=M.(dl)/dt` [`Where,l=l_0sinomegat`]
`E_s=Momegal_0cos(omegat)`
=`1xx2pixx50xx2/picos(2pixx50xxt)`
`because omega=2pin`
For t= 0 we have
`E_s 4xx50 =200V`