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 = 14.0 Ohm, V_rms = 30.0 V for the generator, and f = 380 Hz for the generator. The time constant for the circuit in part a is r = 3 times 10^-4 s. What is the rms current in the circuit in part c?

Explanation / Answer

time constant T = RC

3*10^-4 = 14*C

C = 2.14*10^-5 F

for part(c)

Vrms = Vmax/sqrt2

Irms = Imax/sqrt2

Imax = w*C*Vmax

Irms = w*c*Vmax/sqrt2

Irms = 2*pi*f*C*Vrms

Irms = 2*pi*380*2.14*10^-5*30

Irms =1.533 A <<<<-------answer

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---