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A network of four 10muF capacitors is connected to a 500 V supply, as shown in figure. Determine (a) the equivalent capacitance of the network and (b) the charge on each capcitor. |
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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)=10muF, C.=(10/3)muF`. The network has C. and `C_(n)` connected in parallel. Thus, the equivalent capacitance C of the network is `C=C.+C_(4)=(10/3+10)muF=13.3muF` (b) 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)=500V` Also, `(Q.)/C_(4)=500V` This given for the given value of the capacitances, `Q=500 xx 10/3 xx 10^(-6) =1.7 xx 10^(-3)C and` `Q.=500 xx 10 xx 10^(-6) =5.0 xx 10^(-3)C` |
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