A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The peak current in the circuit is
A) 8.24 A.
B) 8.84 A.
C) 11.0 A.
D) 12.4 A.
E) 23.5 A.

Question

Quizplus · Accepted Answer

The peak current in an AC circuit is given by I_peak = V_peak / Z, where Z is the impedance. For a series RC circuit, Z = sqrt(R^2 + (1/(ωC))^2), where ω = 2πf. Substituting the given values, Z = sqrt(10^2 + (1/(2π*60*300*10^-6))^2) ≈ 13.35 Ω. Thus, I_peak = 110 V / 13.35 Ω ≈ 8.24 A.

A 600-Hz Ac Generator with a Peak Voltage of 110 V