What is the equivalent capacitance between $A$ and $D$ of the network shown in Fig.?

Question

Aicharya · Accepted Answer

The series combination of $C_{2}$ and $C_{3}$ is equivalent to a capacitance $C^{\prime}$ given by $\frac{1}{C^{\prime}}=\frac{1}{C_{2}}+\frac{1}{C_{3}}$ or $C^{\prime}=\frac{C_{2} C_{3}}{C_{2}+C_{3}}=\frac{200 \times 200}{200+200}=100 \mathrm{pF}$ Therefore the circuit reduces to the one shown in Fig. 11.93(a). The equivalent capacitance between points $A$ and $B$ is $C^{\prime \prime}=C_{1}+C^{\prime}=100+100=200 \mathrm{pF}$ The circuit may be further simplified to that in Fig. $11.93(\mathrm{~b})$. The equivalent capacitance $C$ of the entire network, i.e., between points $A$ and $D$, is now that of the series combination of $C^{\prime \prime}$ and $C_{4}$. Thus $\begin{aligned} \frac{1}{C} &=\frac{1}{C^{\prime \prime}}+\frac{1}{C^{4}}=\frac{1}{200}+\frac{1}{100} \ &=\frac{3}{200} \quad \text { or } \quad C=\frac{200}{3} \mathrm{pF} \end{aligned}$ Hence the correct choice is (c).

What is the equivalent capacitance between \(A\) and \(D\) of the network shown in Fig.?

Solution