Question

Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.

Answer 1

R = 300Ω

C = 20μF = 20 × 10^–6F

L = 1H, Z = 500(from 14)

Electric Energy stored in Capacitor = (1/2)CV² = (1/2) × 20 × 10^–6 × 50 × 50 = 25 × 10^–3J = 25mJ

Magnetic field energy stored in the coil = (1/2)LI₀² = (1/2) × 1 × (0.1)² = 5 × 10^–3J = 5mJ