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(B) First,simplify the circuit by combining capacitors in parallel. The $23 \mu F$ and $12 \mu F$ capacitors are in parallel,giving $C_p = 23 + 12 = 3

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(B) First,simplify the circuit by combining capacitors in parallel. The $23 \mu F$ and $12 \mu F$ capacitors are in parallel,giving $C_p = 23 + 12 = 35 \mu F$. Similarly,the two $1 \mu F$ capacitors on the right are in parallel,giving $C_p = 1 + 1 = 2 \mu F$.Now,the circuit forms a Wheatstone bridge structure with capacitors $C_1 = 35 \mu F$,$C_3 = 10 \mu F$,$C_2 = 7 \mu F$,and $C_4 = 2 \mu F$,with a $13 \mu F$ capacitor in the middle.Check the ratio: $\frac{C_1}{C_3} = \frac{35}{10} = 3.5$ and $\frac{C_2}{C_4} = \frac{7}{2} = 3.5$.Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$,the bridge is balanced,and the middle $13 \mu F$ capacitor can be removed.Now,the upper branch has $35 \mu F$ and $7 \mu F$ in series: $C_{up} = \frac{35 	imes 7}{35 + 7} = \frac{245}{42} = \frac{35}{6} \mu F$.The lower branch has $10 \mu F$ and $2 \mu F$ in series: $C_{low} = \frac{10 	imes 2}{10 + 2} = \frac{20}{12} = \frac{10}{6} \mu F$.These two branches are in parallel,so $C_{eq} = \frac{35}{6} + \frac{10}{6} = \frac{45}{6} = 7.5 \mu F$.

Find the equivalent capacitance across $A$ and $B$ in $\mu F$.

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