A `50 Hz` alternating current of peak value `1` ampere flows through the primary coil of a transformer. If the mutual inductance between the primary secondary be `1.5` henry, then the peak value of the induced voltage is

Options

A. `75 V`
B. `150 V`
C. `100 V`
D. None of these

Answer

c

Explanation

`di= 2-(1-1)=2A`
`dt=((1)/(100))`
`M=0.05H`
`e=M(di)/(dt)= 0.5((2)/(1//100))`
`=0.5xx2xx100`
`=100V`