Constants Part A in (Figure 1) G = 8.00 mF , C2 = 7.00 mF . C.-11.0 mF?and C4 11

Question

Constants Part A in (Figure 1) G = 8.00 mF , C2 = 7.00 mF . C.-11.0 mF?and C4 110 mF The capacitor network is connected to an applied potential differnce Va After the charges on the capacitors have reached their final values, the voltage across Ca is 400 V What is the vollage acTss C1? Express your answer with the appropriate units. VValue Units Submit Request Answer Part B Figure 1 of 1 What is the voltage across C2 Express your answer with the appropriate units. C1 C2 Ca VValue Units Submit Request Answer Part C

Explanation / Answer

Part A -

Capacitors C1 and C2 are in series.

So, charge Q1 and Q2 across them shall be equal.

Means, Q1 = Q2

Voltage across C3 = V3 = 40.0 V

For the capacitors, the expression for voltage and Charge is -

Q = C*V

So, for C1 and C2 -

C1*V1 = C2*V2

=> V1 = (C2 / C1)*V2 = (7/8)*V2 = 0.875*V2

and we have -

V1 + V2 = V3 = 40

=> 0.875*V2 + V2 = 40

=> V2 = 40 / (1.875) = 21.3 V

and V1 = 40 - 21.3 = 18.7 V

So, voltage across C1 = 18.7 V

Part B -

Voltage across C2 = V2 = 21.2 V

Part C -

First find out the equivalent capacitance of C1, C2 and C3.

C1 and C2 are in series.

So, its equivalent capacitance = (8x7) / (8+7) = 3.7 mF

This combination is in parallel with C3.

So, the total capacitance = 3.7 + 11 = 14.7 mF

So, charge across this combination = 14.7 x 10^-3 x 40 mC

This is the charge appear across C4 because this combination is in series with C4.

Therefore, voltage across C4 = V4 = (14.7 x 10^-3 x 40) / (11 x 10^-3) = 53.45 V

Part D -

Voltage applied to the network, Vab = 40 + 53.45 = 93.45 V

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