An LC circuit includes a 0.045 µF capacitor and a 390
µH inductor. If the capacitor is rated at 600 V, what is the
maximum inductor current that will keep the capacitor within its
rating?
A
We know that, stored energy of a capacitor is
U = 1/2 C V^2
U = 1/2 x 0.045 x 10^-6 x 600^2 = 0.0081 J
that of an inductor = 1/2 L I^2
1/2 L I^2 = 0.0081
I = sqrt (2 x 0.0081 / L) = sqrt (2 x 0.0081/390 x 10^-6) = 6.45 A
Hence, I = 6.45 A
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