Question

An ac source having a maximum voltage output of 30V is connected in series with a 50 olms resistor, a 0.60 H inductor, and a 20 microF capacitor. What is the maximum charge on the capacitor when the frequency of the source is adjusted so tht the circuit is in resonance? The answers is 2.1mC b but i dont know how to get it.

Answer 1

The resonance frequency can be calculate from the following relation,

ω_c = 1/sqrt(L*C) = 1/sqrt(0.6*20 * 10^-6) = 288.67 rad/s

So the reaction of the capacitor, Xc = 1/(ω_c*C) = 1/(288.67*20*10^-6) = 173.20 ohms

The maximum currect in circuit while resonance is given by, I = Vmax/R = 30/50 = 0.6 A

so the maximum voltage across the capacitor will be the maximum current times the reactance of the capacitor.

Vc = 0.6*173.2 = 103.92 V

The charge is the voltage times the capacitance. So q = C*Vc = 20*10^-6 *103.92 = 0.0020784 C or q = 2.1 mC.