Question

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

• A. $\frac{14C}{7}$
• B. $\frac{22C}{7}$
• C. $\frac{31C}{7}$
• D. $\frac{44C}{7}$

Answer 1

The correct option is B $\frac{22C}{7}$

As we can see the plane of symmetry passes through $\text{ADC}$. So, $\text{B}$ and $\text{D}$ are equipotential points.


Thus, $V_B=V_D$

Upon redrawing the circuit,


Upon further simplication of circuit:

$C^{\prime }=[\frac{\left(C\right)\left(2C\right)}{C+2C}+2C]=\frac{8C}{3}$


Now, equivalent capacitance,

$C_{eq}=[\frac{(\frac{8C}{3}\times 2C)}{\left(\frac{8C}{3}\right)+2C}+2C]=\frac{22C}{7}$

Hence, option (b) is correct answer.