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Q.
The equivalent capacitance for the network shown in the figure is $\frac{k}{7} pF$. Find $k$.
Electrostatic Potential and Capacitance
Solution:
Capacitance of $C_{1}=C_{4}=100\, pF$
Capacitance of $C_{2}=C_{3}=400\, pF$
Supply voltage, $V=400\, V$
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 $C'=200\, pF$
Capacitors $C_{1}$ and $C'$ are in parallel, therefore their equivalent capacitance is
$C''=C'+C_{1}=200+100=300 pF$
Capacitors $C''$ and $C_{4}$ are connected in series, therefore their equivalent capacitance is
$\frac{1}{C_{ eq }}=\frac{1}{C''}+\frac{1}{C_{4}}=\frac{1}{300}+\frac{1}{400}$
$\frac{1}{C_{ eq }}=\frac{7}{1200}$
$\therefore C_{ eq }=\frac{1200}{7} pF$
