Question

Lc circuit consists of a 20.0 mh inductor and a 0.150 µf capacitor. if the maximum instantaneous current is 0.400 a, what is the greatest potential difference across the capacitor?

Answer 1

The following expression is applicable:

Max. inductor energy = Max. capacitor energy

Where;
Max. inductor energy = LI^2/2, with L = 20.0 mH, I = 0.400 A
Max. capacitor energy = CV_max^2/2, C = 0.150 micro Faraday, V_max = Max. potential difference

Substituting;
LI^2/2 = CV^2/2
LI^2 = CV^2
V^2 = (LI^2)/C
V_max = Sqrt [(LI^2)/C] = Sqrt [(20*10^-3*0.4^2)/(0.15*10^-6)] = 146.06 V