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
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 C4μF 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