Part a of the drawing shows a resistor and a charged capacitor wired in series.

Question

Part a of the drawing shows a resistor and a charged capacitor wired in series. When the switch is closed, the capacitor discharges as charge moves from one plate to the other. Part b shows the amount q of charge remaining on each plate of the capacitor as a function of time. In part c of the drawing, the switch has been removed and an ac generator has been inserted into the circuit. The circuit elements in the drawing have the following values: R = 20.0 ohm, V_rms = 18.0 V for the generator, and f = 400 Hz for the generator. The time constant for the circuit in part a is T = 3 times 10^-4 s. What is the rms current in the circuit in part c?

Explanation / Answer

given:
R = 20 ohms
Vrms = 18 v
f = 400 Hz
T = 3*10^-4 s
I rms=?

as we know that time constant T =RC
3*10^-4 = 20*C
C = 1.5*10^-5 F

XC = 1/(2 pi fC)
= 1/(2*3.14*400*1.5*10^-5)
XC = 26.54

Z = sqrt(R^2 + Xc^2) = sqrt(20*20+26.54*26.54)=33.23 ohms

Irms = Vrms / Z = 18 / 33.23 = 0.5416 A (rms)

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