Question

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

Answer 1

Correct Answer - B
`I_(0)=(2)/(pi)A,L=1H,e_(0)=?f=50Hz`
`e_(0)=omegaLI_(0)`
`2pifLI_(0)`
`=2xx3.14xx50xx(2)/(pi)xx1`
=200V.