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

2 μF

6 μF

4 μF

8 μF

Solution

4 μF

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 = 4 μF )

Therefore, no charge flows through C₂.

As C₁ and C₅ are in series.

∴ Their equivalent capacitance,

C' = $\frac{C_{1} \times C_{5}}{C_{1} + C_{5}}$

= $\frac{4 \times 4}{4 + 4}$

C' = 2 μF

Similarly, C₄ and C₃ are in series. Therefore, their equivalent capacitance.

C'' = $\frac{C_{3} \times C_{4}}{C_{3} + C_{4}}$

= $\frac{4 \times 4}{4 + 4}$

C'' = 2 μF

Now, C' and C" are in parallel. Therefore, effective capacitance between A and B

= C' + C"

= 2 + 2

= 4 μF

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