Question

Find equivalent capacitance between $A$ and $B$

• A. $\frac{25}{32}\mu F$
• B. $\frac{32}{25}\mu F$
• C. $\frac{32}{9}\mu F$
• D. $\frac{9}{32}\mu F$

Answer 1

Answer: (B)

Capacitors, $2$ and $3$ are in series, their equivalent capacitance $C^{\prime }$ is given by
$\frac{1}{C^{\prime }}=\frac{1}{1\mu F}+\frac{1}{8\mu F}=\frac{9}{8}\mu F$
or $C^{\prime }=\frac{8}{9}\mu F$
Capacitors 5 and 4 are in series, their equivalent capacitance $C^{\prime \prime }$ is given by
$\frac{1}{C^{\prime \prime }}=\frac{1}{8\mu F}+\frac{1}{4\mu F}=\frac{3}{8}\mu F$
or $C^{\prime \prime }=\frac{8}{3}\mu F$
Now, the capacitance $C$' and $C$'' are in parallel [Figure 1 ] their equivalent capacitance $C$''' will be

$C^{\prime \prime \prime }=\frac{8}{9}\mu F+\frac{8}{3}\mu F=\frac{32}{9}\mu F$
Now capacitance $C$''' and capacitor 1 are in series [Figure 2], hence the equivalent capacitance between A and B is given by

$\frac{1}{C_{eqi}}=\frac{1}{C^m}+\frac{1}{2\mu F}$
$=\frac{9}{32\mu F}+\frac{1}{2\mu F}=\frac{25}{32}\mu F$
or $C_{eqi}=\frac{32}{25}\mu F$