thanks in advance OE-4 F capacitor is charged to a potential of 50.0 V. It is co

Question

thanks in advance

OE-4 F capacitor is charged to a potential of 50.0 V. It is connccted to a 5.0 H inductancc at time t = 0. What is the percent of the energy stored in the magnetic field at t=0.2 s?

Explanation / Answer

we know that q = i t ==> i = q / t and q = C V = 1.0 x 10^-4 F * 50V = 5 x 10^-3 C Total energy U = UB + UE = 0.5 L i^2 + 0.5 q^2 / C = 0.5 L q^2 / t^2 + 0.5 q^2 / C = 0.5 q^2 [ L/t^2 + 1/C] at t = 0 we have U = 0.5 q^2 / C = 0.5 * 25 x 10^-6 / 1.0 x 10^-4 = 0.125 J at t = 0.2 U' = 0.5 * 25 x 10^-6 * [5 / 0.2^2 + 1 / 1.0 x 10^-4] = 0.1265625 J Therefore U / U' = 0.125 / 0.1265625 = 0.9876 = 98.76 %

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