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(B) The given circuit is a Wheatstone bridge configuration. Let the capacitors be $C_1 = 1 \mu F$,$C_2 = 3 \mu F$,$C_3 = 2 \mu F$,$C_4 = 6 \mu F$,and

Vedclass · Accepted Answer

$\frac{9}{4} \mu F$

Vedclass · Answer

(B) The given circuit is a Wheatstone bridge configuration. Let the capacitors be $C_1 = 1 \mu F$,$C_2 = 3 \mu F$,$C_3 = 2 \mu F$,$C_4 = 6 \mu F$,and the central capacitor $C_5 = 5 \mu F$.Check the ratio of capacitors in the arms: $\frac{C_1}{C_3} = \frac{1}{2}$ and $\frac{C_2}{C_4} = \frac{3}{6} = \frac{1}{2}$.Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$,the Wheatstone bridge is balanced.In a balanced Wheatstone bridge,no charge flows through the central capacitor $C_5$,so it can be removed from the circuit.Now,the circuit consists of two parallel branches: one with $C_1$ and $C_2$ in series,and the other with $C_3$ and $C_4$ in series.Equivalent capacitance of the upper branch: $C_{up} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 3}{1 + 3} = \frac{3}{4} \mu F$.Equivalent capacitance of the lower branch: $C_{low} = \frac{C_3 C_4}{C_3 + C_4} = \frac{2 	imes 6}{2 + 6} = \frac{12}{8} = \frac{3}{2} \mu F$.Since these two branches are in parallel,the total equivalent capacitance $C_{AB} = C_{up} + C_{low} = \frac{3}{4} + \frac{3}{2} = \frac{3 + 6}{4} = \frac{9}{4} \mu F$.

Find the equivalent capacitance between the points $A$ and $B$ in the given figure.

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