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

Explanation:The impedance of the circuit is:$$Z = \sqrt{R^2 + (1/2\pi fC)^2}$$$$= \sqrt{(10.0 \Omega)^2 + (1/(2\pi (60.0 Hz)(300 \mu F))^2}$$$$= 11.0 \Omega$$The peak current is:$$I = V/Z = 110 V/11.0 \Omega = 10.0 A$$

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