An ac source having a maximum voltage output of 30 V is connected in series with a 50-Ω resistor,a 0.60-H inductor,and a 20-µF capacitor.What is the maximum charge on the capacitor when the frequency of the source is adjusted so that the circuit is in resonance?
A) 1.0 mC
B) 0.00 mC
C) 2.1 mC
D) 4.2 mC
E) 8.4 mC

Question

Quizplus · Accepted Answer

At resonance, the reactance of the inductor and capacitor will cancel out, leaving only the resistance in the circuit. Thus, the circuit can be simplified to a simple series circuit with a resistance of 50 Ω.

The maximum current that can flow through the circuit is given by:
Imax = Vmax/R = 30 V/50 Ω = 0.6 A

The maximum charge on the capacitor can be found using the formula:
Qmax = C * Vmax
where C is the capacitance of the capacitor and Vmax is the maximum voltage across the capacitor.

At resonance, the maximum voltage across the capacitor will be equal to the maximum voltage output of the source, which is 30 V. Thus:
Qmax = (20 × 10^-6 F) * 30 V = 0.6 mC

Therefore, the best choice for the maximum charge on the capacitor is C) 2.1 mC.

An Ac Source Having a Maximum Voltage Output of 30