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(B) $1$. The two $4 \, \mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 4 \, \mu F + 4 \, \mu F = 8 \, \mu F$.$2$.

Vedclass · Accepted Answer

$\frac{24}{43} \, \mu F$

Vedclass · Answer

(B) $1$. The two $4 \, \mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 4 \, \mu F + 4 \, \mu F = 8 \, \mu F$.$2$. The $2 \, \mu F$ and $3 \, \mu F$ capacitors in the upper branch are in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$,so $C_1 = \frac{6}{5} \, \mu F$.$3$. Similarly,the $2 \, \mu F$ and $3 \, \mu F$ capacitors in the lower branch are in series. Their equivalent capacitance $C_2 = \frac{6}{5} \, \mu F$.$4$. Now,the circuit consists of three capacitors $C_1$,$C_p$,and $C_2$ connected in series between $A$ and $B$.$5$. The total equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_p} + \frac{1}{C_2} = \frac{5}{6} + \frac{1}{8} + \frac{5}{6}$.$6$. $\frac{1}{C_{eq}} = \frac{10}{6} + \frac{1}{8} = \frac{5}{3} + \frac{1}{8} = \frac{40 + 3}{24} = \frac{43}{24}$.$7$. Therefore,$C_{eq} = \frac{24}{43} \, \mu F$.

The effective capacitance between $A$ and $B$ in the given figure is:

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