Question

Four capacitors, $C_1,C_2,C_3$ and $C_4$ are connected as shown in figure below. Calculate equivalent capacitance of the circuit between points X and Y.

Answer 1

$\frac{1}{C_1^{\prime }}=\frac{1}{30}+\frac{1}{20}$ $(\because \frac{1}{C_1^{\prime }}=\frac{1}{C_2}+\frac{1}{C_3})$

$=\frac{2+3}{60}$

$=\frac{5}{60}=\frac{1}{12}$

$C_1^{\prime }=12\mu F$

$C_1^{\prime \prime }=12+48=40\mu F$ $(\because C_1^{\prime \prime }=C_1^{\prime }+C_4)$

$\frac{1}{C}=\frac{1}{10}+\frac{1}{40}$

$=\frac{4+1}{40}$

$\frac{5}{40}=\frac{1}{8}$

$C=8\mu F$