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Question

If the two terminals of a battery of 40 V are connected across A and B, the charge flown through the battery will be

Given: C1=5 μF,C2=2.5 μF,C3=6 μF,C4=2 μF and C5=15 μF




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Solution

Upon redrawing the circuit,


Given: C1=5 μF,C2=2.5 μFC3=6 μF,C4=2 μF and C5=15 μF

C1C4=52=2.5

C5C3=156=2.5

The given arrangement is balanced wheatstone bridge.

Now, C2 capacitor can be neglected such that C1 and C5 are in series and C4 and C3 are in series, and their combination are in parallel.

Ceq=214μF

Q=CeqV=(214×40)μC

Q=210μC

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