Question

A network of four $20\mu F$ capacitors is connected to a $600V$ supply as shown in the figure.

The equivalent capacitance of the network is

• A. $30.26\mu F$
• B. $20\mu F$
• C. $26.67\mu F$
• D. $10\mu F$

Answer 1

Answer: (C)

From the figure, $C_1$, $C_2$, $C_3$ are connected in series
$\therefore \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...\left(i\right)$
$=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{3}{20}\mu F$
or $C_s=\frac{20}{3}\mu F$
Now $C_s$ is in parallel with $C_4$
$\therefore$ equivalent capacitance,
$C_{eq}=C_s+C_4=\frac{20}{3}+20=\frac{80}{3}$
$=26.67\mu F$