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?
1 answer:
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
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