Question

Twelve capacitors each of capacitance $C$ are connected together so that each lies along the edge of the cube as shown in figure. The equivalent capacitance between $1$ and $4$ will be:

• A. $\frac{5C}{7}$
• B. $\frac{4C}{3}$
• C. $\frac{12C}{7}$
• D. $\frac{7C}{4}$

Answer 1

The correct option is C $\frac{12C}{7}$

The points $2$ and $5$ are symmetrically located w.r.t points $1$ and $4$, so they are at the same potential.

Similarly, points $3$ and $8$ are symmetrically located w.r.t points $1$ and $4$, so they are at the same potential.

So, redrawing the circuit,


Reduced form of above circuit will be as follows


For lower branch:

$\frac{1}{(C_{eq}{)}_1}=\frac{1}{2C}+\frac{1}{C}+\frac{1}{2C}=\frac{1+2+1}{2C}$

$\Rightarrow (C_{eq}{)}_1=\frac{2C}{4}=\frac{C}{2}$



The equivalent capacitance for $2C,\frac{5C}{2},2C$ will be,

$\frac{1}{(C_{eq}{)}_2}=\frac{1}{2C}+\frac{2}{5C}+\frac{1}{2C}=\frac{5+4+5}{10C}$

$\Rightarrow (C_{eq}{)}_2=\frac{10C}{14}=\frac{5C}{7}$

Thus net capacitance between $1$ and $4$ will be,

$C_{net}=C+(C_{eq}{)}_2=C+\frac{5C}{7}=\frac{12C}{7}$

Hence, option $\left(c\right)$ is correct.