A 12 V battery is connected to a 6 Ω, 10 H coil, through a closed switch. The switch is suddenly opened in 1 ms. The average induced emf across the coil will be-

20,000V

The correct answer is Option (2) → 20,000V

Given:

Battery voltage = 12 V, Resistance = 6 Ω, Inductance $L = 10 \, H$, Time $\Delta t = 1 \, ms = 1 \times 10^{-3} \, s$

Current in the circuit just before opening:

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

When switch is opened, the current falls from $I = 2 \, A$ to $0 \, A$ in $\Delta t = 10^{-3} \, s$.

Average induced emf:

$E = L \cdot \frac{\Delta I}{\Delta t}$

$E = 10 \cdot \frac{2}{10^{-3}}$

$E = 2 \times 10^{4} \, V$

Answer: $2.0 \times 10^{4}$ V