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 = 30.0 V for the generator, and f = 390 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

Since Time Constant

T=R*C

=>C=T/R =3*10-4/12=2.5*10-5 F

Capacitive reactance

XC=1/2pi*f*C=1/2pi*390*2.5*10-5 =16.32 ohms

Impedance

Z=sqrt[R2+XC2]=sqrt[122+16.322]=20.26 ohms

Rms current

I=V/Z =30/20.26=1.48 A

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects