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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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  • 1 decade ago
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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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