Question

Eight identical capacitor of capacitance $C$ each are connected along edge of a pyramid having square base, $\text{ABCD}$ as shown. Equivalent capacitance across $\text{A}$ and $\text{D}$ is:

• A. $\frac{8C}{15}$
• B. $\frac{22C}{7}$
• C. $\frac{15C}{8}$
• D. $\frac{44C}{7}$

Answer 1

The correct option is C $\frac{15C}{8}$

The circuit can be redrawn as shown below,

Now, using the concept of symmetry we have,

Now, equivalent capacitance along $\text{ABCD}$,

$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{\frac{3C}{2}}+\frac{1}{C}$

$\Rightarrow \frac{1}{C^{\prime }}=\frac{2}{C}+\frac{2}{3C}=\frac{6+2}{3C}$

$\Rightarrow \frac{1}{C^{\prime }}=\frac{8}{3C}$

$\Rightarrow C^{\prime }=\frac{3C}{8}$
So, the equivalent capacitance across $\text{AD}$,

$C_{eq}=\frac{3C}{2}+\frac{3C}{8}=\frac{15C}{8}$

Hence, option (c) is correct answer.