Question

Ex. Find the equivalent capacitance across the point A and B

• A. 1µF
• B. 2µF
• C. 5µF
• D. 4µF

Answer 1

Answer: (D)

4µF

Answer 2

Assuming the capacitor between E and D is $2 \mu F$ instead of $6 \mu F$, the circuit becomes a balanced Wheatstone bridge since the ratio of capacitances in the arms is equal ($C_{EC}/C_{ED} = 6/2 = 3$ and $C_{CF}/C_{DF} = 6/2 = 3$). In a balanced Wheatstone bridge, no charge flows through the diagonal capacitor ($C_{CD}$), so it can be removed. The equivalent capacitance is then the parallel combination of the series combinations of the two pairs of arms.

The series combination of $C_{EC}$ and $C_{CF}$ is $\frac{6 \times 6}{6+6} = 3 \mu F$.

The series combination of $C_{ED}$ and $C_{DF}$ is $\frac{2 \times 2}{2+2} = 1 \mu F$.

The equivalent capacitance across A and B is the parallel combination of these two, which is $3 + 1 = 4 \mu F$.