A current of 10 A in primary of a circuit is reduced to zero at a uniform rate in \( 10^{-3} \)s. If coeff. of mutual inductance is 3 H, the induced emf in the secondary would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3\times10^{4}\,\mathrm{V} \)
The emf induced in the secondary is proportional to the rate of change of current in the primary: \( |\varepsilon| = M\,\frac{dI}{dt} \), where M is the coefficient of mutual inductance. The current falls uniformly, so the rate of change is constant.
\( \displaystyle \left|\frac{dI}{dt}\right| = \frac{10\,\mathrm{A}}{10^{-3}\,\mathrm{s}} = 10^{4}\,\mathrm{A\,s^{-1}} \)
\( |\varepsilon| = 3\,\mathrm{H}\times10^{4}\,\mathrm{A\,s^{-1}} = 3\times10^{4}\,\mathrm{V} \)