First slide
Inductance
Question

An alternating current of frequency 200 rad/sec and peak value 1A as shown in the figure, is applied to the primary of a transformer. If the coefficient of mutual induction between the primary and the secondary is 1.5 H, the voltage induced in the secondary will be

Moderate
Solution

e = - M\frac{{di}}{{dt}} = - 1.5\frac{{(1 - 0)}}{{(T/4)}} = - \frac{6}{T},T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{200}} = \frac{\pi }{{100}}
\Rightarrow \,|e|\, = \frac{{600}}{\pi } = 190.9\,V\tilde --\,191V

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