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=12.0 ?, Vrms = 24.0 V for the generator, and f = 395 Hz for the generator. The time constant for the circuit in part a is ? = 3 x10-4 s. What is the rms current in the circuit in part c?

Explanation / Answer

time constant=R*C

==>3*10^(-4)=12*C

==>C=2.5*10^(-5) F

then capacitive impedance=-j/(2*pi*frequency*C) [note: here j=square root of -1]

=-j/(2*pi*395*2.5*10^(-5))=-j16.117 ohms

hence net impedance of the series RC circuit=12-j16.117 ohms

as rms current=rms voltage/magnitude of impedance

here magnitude of impedance=sqrt(12^2+16.117^2)=20.094 ohms

hence rms current=rms voltage/20.094=24/20.094=1.1944 A

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