A circuit containing a 100 μF capacitor and a 40 Ω resistor connected in series is supplied by a 110 V, 60 Hz AC source. What is the peak current in the circuit?

3.24 A
4.25 A
2.25 A
5.20 A

Correct answer: 3.24 A

Solution

Xc = 1/(2*pi*60*100e-6) = 26.5 ohm. Z = sqrt(40^2 + 26.5^2) = 48 ohm. Irms = 110/48 = 2.29 A, so peak = sqrt(2)*2.29 = 3.24 A.

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