InterviewSolution
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InterviewSolution
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A network of four 10 µF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) |
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Answer» Solution :(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 `1/C. = 1/C_(1) + 1/C_(2) + 1/C_(3)` For `C_(1) = C_(2) = C_(3) = 10 muF, C. = (10//3) muF`.The network has C′ and `C_(4)`connected in parallel. THUS, the equivalent capacitance C of the network is `C = C. + C_(4) =(10/3 + 10) muF = 13.3 muF` (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 `Q/C_(1) + Q/C_(2) + Q/C_(3) = 500 V` Also, `Q.//C_(4) = 500 V` This gives for the given VALUE of the capacitances. `Q = 500 V xx 10/3 muF = 1.7 xx 10^(-3)` C and `Q. = 500 V xx 10 muF = 5.0 xx 10^(-3)` C |
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