Question

Effective capacitance between A and B in the figure shown below is ?

• A. $\frac{3}{14}\mu F$
• B. $\frac{14}{3}\mu F$
• C. 21$\mu F$
• D. 23$\mu F$

Answer 1

Answer: (B)

$\frac{14}{3}\mu F$

Answer 2

The given circuit is a Wheatstone bridge configuration of capacitors. Let the capacitors be denoted as follows:

$C_1 = 3 \mu F$ (between A and the top node) $C_2 = 6 \mu F$ (between the top node and B) $C_3 = 4 \mu F$ (between A and the bottom node) $C_4 = 8 \mu F$ (between the bottom node and B) $C_5 = 2 \mu F$ (between the top node and the bottom node)

In a Wheatstone bridge with capacitors, the bridge is balanced if the ratio of capacitances in adjacent arms is equal, i.e., $\frac{C_1}{C_3} = \frac{C_2}{C_4}$.

Let's check the condition for the given values: $\frac{C_1}{C_3} = \frac{3}{4}$ $\frac{C_2}{C_4} = \frac{6}{8} = \frac{3}{4}$

Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$, the Wheatstone bridge is balanced.

When the bridge is balanced, the potential difference across the capacitor in the middle arm ($C_5$) is zero. Therefore, no charge flows through or accumulates on the capacitor $C_5$, and it can be removed from the circuit without affecting the effective capacitance between A and B.

After removing $C_5$, the circuit simplifies to two parallel branches. The top branch consists of $C_1$ and $C_2$ in series, and the bottom branch consists of $C_3$ and $C_4$ in series. These two series combinations are connected in parallel between A and B.

The equivalent capacitance of $C_1$ and $C_2$ in series is $C_{series1} = \frac{C_1 C_2}{C_1 + C_2}$. $C_{series1} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \mu F$.

The equivalent capacitance of $C_3$ and $C_4$ in series is $C_{series2} = \frac{C_3 C_4}{C_3 + C_4}$. $C_{series2} = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \mu F$.

The effective capacitance between A and B is the parallel combination of $C_{series1}$ and $C_{series2}$. $C_{AB} = C_{series1} + C_{series2}$. $C_{AB} = 2 + \frac{8}{3} = \frac{6}{3} + \frac{8}{3} = \frac{6+8}{3} = \frac{14}{3} \mu F$.

The effective capacitance between A and B is $\frac{14}{3} \mu F$.