Discussion Forum

An alternating current of peak value ( $\frac {2}{\pi}$) ampere flows through the primary coil of the transformer. The coefficient of mutual  inductance  between primary and secondary  coil is 1H, the peak emf induced in the secondary coil is (Frequency of Ac= 50 Hz)

Answer:

Explanation:

according to question , peak value of current

$l_{0}=\sqrt{2}\times l_{rms}=\frac{2}{\pi}A$

Coefficient of mutual inductance =1H

as, we know  induced emf in secondary coil is given by

  $E_{s}=M.\frac{dl}{dt}$       [ where , l= l₀ sinωt]

 $E_{s}=M.\omega l_{0}\cos (\omega t)$

  = $1\times 2\pi\times50\times\frac{2}{\pi}\cos(2\pi\times50\times t)$

                                               $(\therefore\omega=2\pi n)$

 For t=0, we have

 E_e = 4 x 50=200 V