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Obtain equivalent capacitance of the following net work, Fig. For a 300V supply determine the charge and voltage across each capacitor.

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The equivalent circuits of network in Fig, are shown in Fig.
In fig as `100 pF and 200 pF` capacitors are in series, pot diff, across `C_(4)` is the ratio `2:1`.
`:. V_(4) = (300x2)/(2+1) = 200V`

`q_(4) = C_(4) V_(4) = (100xx10^(-12)) 200 = 2xx10^(-8) C`
pot diff. across `200 pF = 300 - 200 = 100V`
This must be the pot.diff across `C_(1) i.e, V_(1) = 100V :. q_(1) = C_(1) V_(1) = (100xx10^(-12))xx100 = 10^(-8) C`
Also pot. diff across `C_(2), C_(3)` in series = 100V `:.V_(2) = V_(3) = (100)/(2) = 50V`
`q_(2) = C_(2) V_(2) = (200xx10^(-12)) xx 50 = 10^(-8)C, q_(3) = C_(3) V_(3) = (200xx10^(-12))xx50 = 10^(-8) C`
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