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.

Question

Navjyot Kalra · Answer

Sol R = 300Ω C = 20 μF = 20 * 10^-6 F L = 1H, Z = 500 (from 14) Ε base 0 = 50 V, I base 0 = E base 0/Z = 50/500 = 0.1 A Electric Energy stored in Capacitor = (1/2) CV^2 = (1/2) × 20 × 10^–6 × 50 × 50 = 25 × 10^–3 J = 25 mJ Magnetic field energy stored in the coil = (1/2) L I base 0^2 = (1/2) * 1 * (0.1)^2 = 5 * 10^-3 J = 5 mJ

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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.

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.

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