12 V battery connected to a coil of resistance 6 Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is :

Solution

Physics sir · 3 steps

Find the steady current first, then see how fast it falls to zero; the 20 V emf gives L directly.

1. Steady current. While the switch is closed the current is constant, so the coil acts like a plain resistor. Then I = 12/6 = 2 A.
2. Change in current. Opening the switch takes the current from 2 A to 0 in Δ t = 1 ms = 10⁻³ s. So (Δ I)/(Δ t) = 2/10⁻³ = 2000 A s⁻¹.

   |ε| = L (Δ I)/(Δ t) ⇒ L = 20/2000 = 10⁻² H

3. So. L = 10⁻² H = 10 mH, which is option (c).

The idea to keep. Self-induced emf depends on how fast the current changes: ε = L Δ I/Δ t. Get the current from Ohm's law first, and convert milliseconds to seconds before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.