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The figure shows a network of five capacitors connectedto a 100 V supply, Calculate the total energy stored in the network. |
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Answer» Solution :The network of capacitors can be redrawn as shown here. Obviously CAPACITOR `C_3 of 2 muF` is short circuited and is useless. Capacitance `C_12` of PARALLEL grouping of `C_4 and C_2` is `C_12 = C_1+ C_2=3 + 3 = 6 muF` Capacitance `C_45` of parallel grouping `C_4 and C_5` is `C_(45) = C_4+ C_5 = 3 + 3 = 6 muF` In the network `C_12 and C_45` are in series. So the net capacitance of the ARRANGEMENT `C = (C_12 xx C45)/((C_12 + C45)) = (6 xx3)/((6 xx 3)) = muF = 2muF` `:.` TOTAL energy stored int he network `u =1/2 CV^2 =1/2xx(2muF) xx(100V)^2` `=1/2 xx 2 xx 10^(-6) xx(100)^2 = 10^(-2) J = 0.01J` 
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