In the arrangement shown in figure, the capacitors are indentical. When S1 is closed and S2 is open, equivalent capacitance between A and B is 4µF. What will be the equivalent capacitor between A and B when S1 is open and S2 is closed?

In the first case, CAB=2C×C2C+C=4⇒C=6µFIn the second case, C'AB=C+C2=32C=32×6µF=9µF

In the arrangement shown in figure, the capacitors are indentical. When S1 is closed and S2 is open, equivalent capacitance between A and B is 4µF. What will be the equivalent capacitor between A and B when S1 is open and S2 is closed?

In the first case, CAB=2C×C2C+C=4⇒C=6µFIn the second case, C'AB=C+C2=32C=32×6µF=9µF

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In the arrangement shown in figure, the capacitors are indentical. When S₁ is closed and S₂ is open, equivalent capacitance between A and B is 4µF. What will be the equivalent capacitor between A and B when S₁ is open and S₂ is closed?

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9 µF

6 µF

12 µF

4.5 µF

answer is A.

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Detailed Solution

In the first case, $C_{AB} = \frac{2 C \times C}{2 C + C} = 4 \Rightarrow C = 6 µ F$

In the second case, $C'_{AB} = C + \frac{C}{2} = \frac{3}{2} C = \frac{3}{2} \times 6 µ F = 9 µ F$

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In the arrangement shown in figure, the capacitors are indentical. When S1 is closed and S2 is open, equivalent capacitance between A and B is 4µF. What will be the equivalent capacitor between A and B when S1 is open and S2 is closed?

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In the arrangement shown in figure, the capacitors are indentical. When S₁ is closed and S₂ is open, equivalent capacitance between A and B is 4µF. What will be the equivalent capacitor between A and B when S₁ is open and S₂ is closed?