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Question
The figure shows a network of five capacitors connected to a 100 V supply. Calculate the total energy stored in the network.
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Solution
As the both 3 uF capacitors are connected in parallel, so net capacitance between branch EH = 3 + 3 = 6μF
Similarly, the capacitance 2 uF and 1uF at corner B are also connected in parallel, so the net capacitance of branch FG = 2 + 1 = 3 μF
If reconstruct the given figure according to the above calculations, we can see that 6 μF capacitors and 3 μF capacitors are connected in series and another 2 μF capacitor is connected in parallel with both of them.
Hence net capacitance Between D and C `= 2 + (3xx6)/(3+6) = 2+2 = 4 mu"F"`
The total capacitance of the circuit, Cnet = 4μF
Total voltage applied, V = 100V
Energy stored in the network `= 1/2 "C"_"net""V"^2 = 1/2 xx 4xx10^-6 xx (100)^2 = 0.02 "J"`
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