Physics - Modern Physics Question with Solution | TestHub

Let a total current of 6 I enter at A. It divides into three equal parts, each of $2\mathrm{I}$, along $\mathrm{AE}$, $\mathrm{AB}$ and $\mathrm{AD}$. At $\mathrm{E},\mathrm{B}$ and $\mathrm{D}$ each the current 2I divides into two equal parts, each of $\mathrm{I}$, along $\mathrm{EF},\mathrm{EH},\mathrm{BF},\mathrm{BC},\mathrm{DH}$ and $\mathrm{DC}$. At $\mathrm{F},\mathrm{H}$ and $\mathrm{C}$, the two currents each of $\mathrm{I}$, combine together to give a current of $2\mathrm{I}$ at each corner. Thus, at $\mathrm{G}$ we get the same current 6 I as shown in the figure.
Let $\mathrm{r}$ be the value of resistance of each $\mathrm{arm}$ of the cube and $\mathrm{R}$ be the joint resistance across the conners A and G. Applying Kirchhoff's law along the loop AEFGA, we get $2\mathrm{Ir}+\mathrm{Ir}+2\mathrm{Ir}=\mathrm{E}$ or $5\mathrm{Ir}=\mathrm{E}$ ...(1) Also, by Ohm's law, $\begin{array}{l}6\mathrm{I}\times \mathrm{R}=\mathrm{E}\\ \Rightarrow 6\mathrm{IR}=5\mathrm{Ir},\text{ using }(1)\end{array}$ or, $\mathrm{R}=\frac{5}{6}\mathrm{r}$. Here, $r=6\Omega$ $\therefore \mathrm{R}=\frac{5}{6}\times 6=5\Omega$