Question

The equivalent capacitance for the network shown in the figure is.

• A. $\frac{1200}{7}pF$
• B. $\frac{1000}{4}pF$
• C. $\frac{1800}{7}pF$
• D. $\frac{1300}{3}PF$

Answer 1

Answer: (A)

$\frac{1200}{7}pF$

Answer 2

:

Capacitance of $C_{1} = C_{4} = 100pF$

Capacitance of $C_{2} = C_{3} = 400pF$

Supply voltage,$V = 400V$

Capacitors $C_{2}$and $C_{3}$are connected in series,

Equivalent capacitance

$C' = \frac{1}{400} + \frac{1}{400} = \frac{2}{400}$ or $C' = 200pF$

Capacitors $C_{1}$and $C$are in parallel their equivalent capacitance

$C'' = C' + C_{1} = 200 + 100 = 300pF$

Capacitors $C''$and $C_{4}$are connected in series

Equivalent capacitance,$\frac{1}{C_{eq}} = \frac{1}{C''} + \frac{1}{C_{4}} = \frac{1}{300} + \frac{1}{400}$

$\frac{1}{C_{eq}} = \frac{7}{1200}$

$\therefore C_{eq} = \frac{1200}{7}pF$