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Grade 11Electric Current

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.

Profile image of Radhika Batra
12 Years agoGrade 11
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
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