A \( 12\,\mathrm{V} \) battery connected to a \( 6\,\Omega \), \( 10\,\mathrm{mH} \) coil through a switch drives a constant current in the circuit. The switch is suddenly opened. Assuming that it took \( 1\,\mathrm{ms} \) to open the switch, the average emf induced across the coil would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 20\,\mathrm{V} \)
In the steady state the coil acts as a plain resistor, so the steady current is found from Ohm's law. When the switch opens, the current falls to zero and the coil produces a self-induced emf that opposes this change.
\( \displaystyle I = \frac{V}{R} = \frac{12}{6} = 2\,\mathrm{A} \)
\( \displaystyle \varepsilon = L\,\frac{|\Delta I|}{\Delta t} = \frac{(10\times10^{-3}) \times 2}{1\times10^{-3}} = 20\,\mathrm{V} \)