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)

$26.67\mu F$

Answer 2

: From the figure, $C_{1,}C_{2,}C_{3}$are connected in series.

$\therefore\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$ …(i)

$= \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{3}{20}\mu F$

or $C_{s} = \frac{20}{3}\mu F$

Now $C_{S}$is in parallel with $C_{4}$

$\therefore$equivalent capacitance,

$C_{eq} = C_{s} + C_{4} = \frac{20}{3} + 20 = \frac{80}{3} = 26.67\mu F$