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Alternating current of peak value ((2)/(...

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

Text Solution

Verified by Experts

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