A 13.5-V battery is connected to terminals A and B in the figure below. (a) Give

Question

A 13.5-V battery is connected to terminals A and B in the figure below. (a) Given that C = 16.6 µF, find the charge on each of the capacitors.

Q7.22 µF = Correct: Your answer is correct. µC

Q16.6 µF = Correct: Your answer is correct. µC

Q4.25 µF = Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. µC

Q12.0 µF = Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. µC

Q8.35 µF = µC

Explanation / Answer

First step find the Ceq

12 uF and 8.35 uF are in series

Ca = C1*C2/(C1+ C2) for series comb.

Ca = 12*8.35/20.35 = 4.923 uF

Now 4.923 uF and 4.25 uF are in parallel comb.

Cb = C1 + C2 for parallel comb.

Cb = 4.923 + 4.250 = 9.173 uF

Now 9.173 uF and 16.6 uF are in series

Cc = 16.6*9.173/(16.6+9.173) = 5.908 uF
Now 5.908 uF and 7.22 uF are in parallel

Ceq = 7.22 + 5.908 = 13.128 uF

Veq = 13.5 V

Q = CV

total charge in circuit

Q = 13.128*13.5*10^-6 = 1.77*10^-4 C

A. In parallel comb, Voltage will be same

V in 7.22 uF = V in 5.908 uF = 13.5 V

Current in 7.22 uF Capacitor

V1 = 13.5 V

C1 = 7.22 uF

Q1 = 13.5*7.22*10^-6 = 0.974*10^-4 C

B.

V in 5.908 uF = 13.5 V

Q in 5.908 uF = (1.77 - 0.974)*10^-4 = 0.796*10^-4 C

So charge in 16.6 uF = 0.796*10^-4 C

Voltage in 16.6 uF = 0.796*10^-4/(16.6*10^-6) = 4.795 V

C.

Voltage in 9.173 uF = 13.5 - 4.795 = 8.705 V

Voltage in 4.25 uF = 8.705 V

Charge in 4.25 uF = 4.25*10^-6*8.705 = 0.369*10^-4 C

D. Voltage in 4.923 uF = 8.705 V

charge in 4.923 uF = (0.796 - 0.369)*10^-4 = 0.427*10^-4

Now 12 uF and 8.35 uF are in series so charge will be same

Charge in 12 uF = 0.427*10^-4 C

Charge in 8.35 uF = 0.427*10^-4 C

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