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Q.
The equivalent capacitance for the network shown in the figure is
Electrostatic Potential and Capacitance
Solution:
Capacitance of $C_{1} = C_{4} = 100 \, pF$
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
Equivalent capacitance
$C^{\prime}=\frac{1}{400}+\frac{1}{400}=\frac{2}{400}$
or $C ^{\prime}=200 pF$
Capacitors $C_{1} C^{\prime}$ are in parallel
Their equivalent capacitance
$C ^{\prime \prime}= C ^{\prime}+ C _{1}=200+100=300 pF$
Capacitors $C ^{\prime \prime}$ and $C _{4}$ are connected in series
Equivalent capacitance $\frac{1}{ C _{ eq }}=\frac{1}{ C ^{\prime \prime}}+\frac{1}{ C _{4}}=\frac{1}{300}+\frac{1}{100}$
$\frac{1}{ C _{ eq }}=\frac{4}{300}$
$\therefore C _{ eq }=\frac{300}{4} pF$
