(a) A 32 resistor is connected in series with a 8 µF capacitor and a battery. Wh

Question

(a) A 32 resistor is connected in series with a 8 µF capacitor and a battery. What is the maximum charge to which this capacitor can be charged when the battery voltage is 9 V? (When entering units, use micro for the metric system prefix µ.)


(b) Initially the capacitor has zero charge. At what time during the charging process will the charge on this capacitor be 45.5 µC?


(c) The capacitor is now fully charged. The battery is removed and the capacitor is discharged through a bulb. What will be the charge on the capacitor after 0.256 ms? (Assume that the bulb has the same resistance as the original resistor.)

Explanation / Answer

part a )

Qmax = C*V

Qo = 8 x 10^-6 F * 9 = 72 x 10^-6 C = 72 uC

part b )

for charging capacitor

Q = Qo(1-e^-t/T)

T = time constant = RC = 32 * 8 x 10^-6 = 2.56 x 10^-4 s

45.5 uc = 72 uC(1-e^-t/2.56 x 10^-4)

e^-(t/2.56x10^-4) = 1-(45.5/72)

e^-(t/2.56x10^-4) = 0.368

take natural log both side

-t/2.56x10^-4 = -0.99952

t = 2.5588 x 10^-4 s ( almost 1 time constant)

part c )

discharging capacitor

Q = Qo(e^-t/RC)

Q = 26.487 uC

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.