A 12 V battery connected to a 6 Ω, 10 mH coil through a switch drives a constant current in the circuit. The switch is suddenly opened. Assuming that it took 1 ms to open the switch, the average emf induced across the coil would be

20 V

The correct answer is Option (2) → 20 V

Given:

Battery voltage $V = 12\text{ V}$

Resistance $R = 6\,\Omega$

Inductance $L = 10\,\text{mH} = 10 \times 10^{-3}\,\text{H}$

Time to open switch $\Delta t = 1\,\text{ms} = 1 \times 10^{-3}\,\text{s}$

Steady current before opening the switch:

$I = \frac{V}{R} = \frac{12}{6} = 2\,\text{A}$

When the switch opens, current drops from $2\,\text{A}$ to $0$ in $1\,\text{ms}$.

Average induced emf across the coil:

$\text{emf} = L \frac{\Delta I}{\Delta t} = (10 \times 10^{-3}) \times \frac{2}{1 \times 10^{-3}} = 20\,\text{V}$

Answer: 20 V