Capacitors C1= 13 uF and C2=25 u F are each charged to 15 V , then disconnected

Question

Capacitors C1= 13 uF and C2=25 u F are each charged to 15 V , then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa.

Part A. Afterward, what is the charge on each capacitor?
Express your answer using two significant figures. Enter your answers numerically separated by a comma

Part B What is the potential difference across each capacitor?
Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Explanation / Answer

While connecting to the battery, both capacitors are equally charged (serial connection), assumed the charge is Q. Q/C1 +Q/C2 = U, we have Q = U / (1/C1+1/C2) = 18/(1/4+1/12) = 54 µCoulomb After they are disconnected to the battery and reconnected to each other, the potential difference across both of them are the same (parallel connection), assume the potential is Un. And we have charge conservation: C1 Un + C2 Un = Q + Q Un = 2 Q / (C1+C2) = 2 *54/(4+12) = 6.75 V initial energy stored in capacitors Ei = 0.5 (Q^2/C1+ Q^2/C2) = 0.5 Q^2 (1/C1+ 1/C2) = 0.5*54^2*(1/4+1/12) = 486 µJ final energy stored in capacitors Ef = 0.5 (C1 U^2 + C2 U^2) = 0.5 U^2 (C1+ C2) = 0.5*6.75^2*(4+12) = 364.5 µJ

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