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In the given network, `C_(1), C_(2)` and `C_(3)` are connected in series. The effective capacitive 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_(3) = 10 mu F. C' = (10//3) nu F`. The network has C' and `C_(4)` connected in parallel. Thus, the equivalent capacitance C of the network is <br> `C = C' + C_(4) = ((10)/(3) + 10) mu F = 13.3 nu 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`. The gives for the given 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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