In a series R-L circuit, battery voltage is 10 V. Resistance and inductance are 10Ω and 10 mH respectively. Find energy stored in the inductor when current reaches 1/e times of maximum value.

Solution:

Given:

• Voltage, V = 10 V
• Resistance, R = 10 Ω
• Inductance, L = 10 mH = 0.01 H

Maximum current in RL circuit:

I_max = V/R = 10/10 = 1 A

Current at the given instant:

I = I_max/e = 1/e A

Energy stored in inductor:

U = (1/2)LI²

U = (1/2) × 0.01 × (1/e)²

U = 0.005/e²

e² ≈ 7.39

U ≈ 0.005/7.39

U ≈ 0.000676 J

U ≈ 0.67 mJ

Answer: 0.67 mJ