Question

The equivalent capacitance for the network shown in the figure is

• A. $\frac{300}{4}pF$
• B. $\frac{1000}{4}pF$
• C. $\frac{1800}{7}pF$
• D. $\frac{1300}{3}pF$

Answer 1

Answer: (A)

Capacitance of $C_1=C_4=100pF$
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
$C^{\prime }=\frac{1}{400}+\frac{1}{400}=\frac{2}{400}$
or $C^{\prime }=200pF$
Capacitors $C_1{C}^{\prime }$ are in parallel
Their equivalent capacitance
$C^{\prime \prime }=C^{\prime }+C_1=200+100=300pF$
Capacitors $C^{\prime \prime }$ and $C_4$ are connected in series
Equivalent capacitance $\frac{1}{C_{eq}}=\frac{1}{C^{\prime \prime }}+\frac{1}{C_4}=\frac{1}{300}+\frac{1}{100}$
$\frac{1}{C_{eq}}=\frac{4}{300}$
$\therefore C_{eq}=\frac{300}{4}pF$