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# For the configuration Fig. 10.116, determine the voltage across each capacitor and charge on each capacitor.?

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All C's and Q's will be in µF or µC

9 and 72 combine for 9*72/81 = 8 (C34)

that in parallel with 10 is 18 (C234)

that in series with 9 is 9*18/27 = 6 (Ct)

So total capacitance is 6 and charge on that is

Q = Cv = 6x24 = 144

Since they are in series, C1 and C234 have this same charge of 144

Voltage across C234 is V = Q/C = 144/18 = 8 volts

Voltage across C1 is 24–8 = 18

This means Q2 = C2*8 = 80

Q34 = C34*8 = 8*8 = 64

Since they are in series, C3 and C4 have this same charge.

V3 = Q3/C3 = 64/9 = 7.11

V4 = Q4/C4 = 64/72 = 0.89

summary

C ... Q ... V

1 .. 144 .. 18

2 .... 80 .... 8

3 .... 64 .. 7.11

4 .... 64 .. 0.89

.

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