Question

The equivalent capacitance for the network shown in the figure is $\frac{k}{7}pF$. Find $k$.

Answer 1

Answer: 1200

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, therefore their equivalent capacitance is
$C^{\prime }=\frac{1}{400}+\frac{1}{400}=\frac{2}{400}$ or $C^{\prime }=200pF$
Capacitors $C_1$ and $C^{\prime }$ are in parallel, therefore their equivalent capacitance is
$C^{\prime \prime }=C^{\prime }+C_1=200+100=300pF$
Capacitors $C^{\prime \prime }$ and $C_4$ are connected in series, therefore their equivalent capacitance is
$\frac{1}{C_{eq}}=\frac{1}{C^{\prime \prime }}+\frac{1}{C_4}=\frac{1}{300}+\frac{1}{400}$
$\frac{1}{C_{eq}}=\frac{7}{1200}$
$\therefore C_{eq}=\frac{1200}{7}pF$