Question

Twelve capcitors each having a capacitance $C$, are connected to form a cube. Find the equivalent capacitance of the cube along its body diagonal? [Take $C=15\mu \text{F}$]

• A. $18\mu \text{F}$
• B. $12\mu \text{F}$
• C. $\frac{6}{5}\mu \text{F}$
  
• D. $\frac{5}{6}\mu \text{F}$
  

Answer 1

The correct option is A $18\mu \text{F}$

Suppose the points $\text{A}$ and $\text{B}$ are connected to a battery.

Suppose the positive termainal of the battery supplies a charge is divided on the three plates connected to $\text{A}$.

By applying symmetry and charge conservation the charge on each capacitor is obtained as shown in the figure.



Potential difference between $\text{AG}$ is given by

$V_{AG}=V_{AE}+V_{CH}+V_{HG}=V$

If $C_{eqv}$ is equivalent capacitance across the diagonal $\text{AG}$, then

$\frac{Q}{C_{{}_{eqv}}}=\frac{\frac{Q}{3}}{C}+\frac{\frac{Q}{6}}{C}+\frac{\frac{Q}{3}}{C}$

$\Rightarrow \frac{1}{C_{eqv}}=\frac{2+1+2}{6C}$

$\Rightarrow C_{eqv}=\frac{6}{5}C$

As, $C=15\mu \text{F}$,

$\Rightarrow C_{eqv}=\frac{6}{5}\times 15=18\mu \text{F}$

Hence, option (a) is correct answer.

Why this question: About body diagonal, all the paths are symmetrical. Thus using charge distribution we can get the equivalent capacitance between the body diagonal.