Question

A network of four $20 \,\mu F$ capacitors is connected to a $600\, V $ 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,
$C_{eq}=C_{s}+C_{4}=\frac{20}{3}+20=\frac{80}{3}$
$=26.67\,\mu F$