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(a) The capacitors `C_(1) , C_(2)` and `C_(3)` are in series <br> `therefore " " (1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(15) + (1)/(15) + (1)/(15)` <br> `(1)/(C_(s)) = (3)/(15) implies C_(S) = 5 mu F `. <br> Now `C_(S)` and `C_(4)`are in parallel combination . Therefore , the equivalent capacitance of the network is <br> `C = C_(s) + C_(4) = 5 + 15 = 20 mu F`. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SB_PHY_XII_10_OD_E02_005_S01.png" width="80%"> <br> (b) Charge on capacitor `C_(4)` is = `C_(4) V = 15 xx 10^(-6) xx 500 = 7500 xx 10^(-6) C = 7500 mu C` <br> Charge on each capacitor , `C_(1) , C_(2)` and `C_(3)` is q = `C_(s) V = 5 xx 10^(-6) xx 500` <br> `= 2500 xx 10^(-6) C = 2500 mu C`Related Video

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