A 12 V battery is connected to a \( 6\,\Omega \), 10 H coil, through a closed switch. The switch is suddenly opened in 1 ms. The average induced emf across the coil will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20,000 V
When the switch is opened, the steady current in the coil falls to zero in 1 ms. By self-induction, the average induced emf is \( L \) times the average rate of change of current.
\( \displaystyle I = \frac{E}{R} = \frac{12\,\mathrm{V}}{6\,\Omega} = 2\,\mathrm{A} \)
\( \Delta I = 2 - 0 = 2\,\mathrm{A} \)
\( \Delta t = 1\,\mathrm{ms} = 1\times10^{-3}\,\mathrm{s} \)
\( \displaystyle |e| = L\,\frac{\Delta I}{\Delta t} = 10\,\mathrm{H} \times \frac{2\,\mathrm{A}}{1\times10^{-3}\,\mathrm{s}} = 2\times10^{4}\,\mathrm{V} \)
So the average induced emf is 20,000 V.