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(a) In the given network, `C_(1), C_(2)` and `C_(3)` are connected in series. The effective capacitance C′ of these three capacitors is given by <br> `(1)/(C') = (1)/(C_(1)) + (1)/(C_(2))+(1)/(C_(3))` <br> For `C_(1) = C_(2) = C_(2) = C_(3) = 10 mu F, C ′ = (10//3) mu F`. The network has C ′ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is <br> `C = C' + C_(4) = ((10)/(3)+10) mu F = 13.3 mu F` <br> (b) Clearly, from the figure, the charge on each of the capacitors, `C_(1), C_(2)` and `C_(3)` is the same, say Q. Let the charge on `C_(4)` be Q′ . Now, since the potential difference across AB is `Q//C_(1)`, across BC is `Q//C_(2)`, across CD is `Q//C_(3)`, we have <br> `(Q)/(C_(1)) + (Q)/(C_(2)) + (Q)/(C_(3)) = 500 V` <br> Also, `Q'//C_(4) =500 V`. <br> This gives for the value of the capacitances, <br> `Q = 500 V xx (10)/(3)mu F =1.7 xx 10^(-3)C` and <br> `Q' = 500 V xx 10 mu F = 5.0 xx 10^(-3)C`Related Video

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