Question

A charged capacitor with C = 730 μF is connected in parallel to an inductor that has L = 0.470 H and negligible resistance. At an instant when the current in the inductor is i = 2.30 A , the current is increasing at a rate of di/dt=89.0A/s.During the current oscillations, what is the maximum voltage across the capacitor?

Answer 1

Voltage across capacitor = Voltage across inductor

Q/C = LdI/dt

Q=LC(dI/dt)=(0.47)(730*10^-6)(89) =0.0305359 C

Total energy stored in capacitor

E=(1/2)LI²+(1/2)(Q²/2C)

Since E =(1/2)CV_max²

=>(1/2)CV_max² = (1/2)LI²+(1/2)(Q²/C)

(730*10^-6)V_max² =0.47*2.3²+(0.0305359²/730*10^-6)

V_max =71.8 Volts