Three capacitors with capacitances C_1 = 6.00 mu F, C_2 = 3.00 mu F, C_3 = 5.00

Question

Three capacitors with capacitances C_1 = 6.00 mu F, C_2 = 3.00 mu F, C_3 = 5.00 mu F. The capacitor network is connected to an applied potential V_ab. After the charges on the capacitors have reached their final values, the charge Q_2 on the second capacitor is 40.0 mu C. What is the charge Q_1 on capacitor C_1? Q_1 = mu C What is the charge on capacitor C_3? Express your answer in microcoulombs to three significant figures. Q_3 = mu C What is the applied voltage, V_ab? Express your answer in volts to three significant figures.

Explanation / Answer

C1 and C2 are parallel

potential across C1 = potential across C2

V1 = V2 = Q1/C1

charge on C2 , Q2 = C2*V1 = C2*Q1/C1 = (C2/C1)*Q1 = (3/6)*40 = 20 uC

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part(B)

charge on Q3 = Q1 + Q2 = 60 uC

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Vab = Vad + Vbd

Vab = (Q1/C1) + (Q3/C3)

Vab = (40/6) + (60/5)

Vab = 18.7 V <<<------answer

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