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. 225 V
D. 300 V

Answer

D

Explanation

Time period of AC
`T=(1)/(n)=(1)/(50)s`
Time interval `Deltat` for current to decrease from 1 A to zero `=(T)/(4)`
`:." "Deltat=(T)/(4)=(1)/(50)xx(1)/(4)=(1)/(200)s`
Change in current, `Deltal=0-1=-1A`
Mean induced emf, `e=-M((Deltal)/(Deltat))`
`=-1.5xx(-1)/(1//200)=300" V"`