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Question

In the given figure, the capacitors C1,C3,C4,C5 have a capacitance 4μF each. If the capacitor C2 has a capacitance 10μF, then effective capacitance between A and B will be
639371_79207889d87b433da8f0ec8e8337c978.png

A
2μF
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B
6μF
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C
4μF
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D
8μF
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Solution

The correct option is C 4μF
When a battery is applied across A and B, then the points b and c will be at the same potential (C1=C4=C3=C5=4μF)
Therefore, no charge flows through C2.
As, C1 and C5 are in series.
Their equivalent capacitance,
C=C1×C5C1+C5=4×44+4=2μF
Similarly, C4 and C3 are in series. Therefore, their equivalent capacitance
C=C3×C4C3+C4=4×44+4=2μF
Now, C and C are in parallel. Therefore, effective capacitance between A and B
=C+C=2+2=4μF

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