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Find the equivalent capacitance of the network shown in the fig , when each capacitor is of 1 mu F . When the ends X and Y are connected to a 6 V battery , find out (i) the charge and (ii) the energy stored in the network |
Answer» Solution :The NETWORK may be REDRAWN as shown in fig  Here , `C_(1) = C_(2) = C_(3) = C_(4) = C_(5) = 1 MU F ` As `(C_(1))/(C_(2)) = (C_(3))/(C_(4))` , hence the network is like a balanced wheatstone bridge where `V_(B) = V_(D)` , Hence capacitor `C_5` may be omitted from the circuit . Now capacitance of series combination of `C_1` and `C_2` is `C_12 = (C_(1) C_(2))/(C_(1) + C_(2)) = (1 xx 1)/(1 +1) = 0.5 mu F ` Similarly capacitance of series combination of `C_3` and `C_4` is `C_34 = 0.5 mu F` SINCE `C_(12)` and `C_(34)` are in parallel across points X and Y . hence the equivalent capacitance of the network `C = C_12 + C_34 = 0.5 + 0.5 = 1.0 mu F ` (i) `therefore` Net charge in the network Q = CV = `(1.0 mu F) xx (6 V) = 6.0 mu C ` (ii) The energy stored in the network `U = (1)/(2) QV = (1)/(2) xx (6.0 mu C) xx 6 V = 18 mu J` |
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