If the resistance of each edge of a cube-shap | vedclass

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(A) To find the equivalent resistance between points $1$ and $7$ (which are diagonally opposite corners of the cube),we consider a total current $I$ e

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$\frac{5 R}{6}$

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(A) To find the equivalent resistance between points $1$ and $7$ (which are diagonally opposite corners of the cube),we consider a total current $I$ entering at point $1$ and leaving at point $7$.By symmetry,the current $I$ splits into three equal parts of $I/3$ at point $1$,flowing through the three edges connected to it.At the next set of nodes,these currents further split. Following the path along any edge,the current distribution is as shown in the figure.Applying Kirchhoff's loop rule along a path from $1$ to $7$ (e.g.,$1 \rightarrow 4 \rightarrow 8 \rightarrow 7$):$V = V_1 - V_7 = I_1 R_1 + I_2 R_2 + I_3 R_3$$V = (I/3)R + (I/6)R + (I/3)R$$V = I R (1/3 + 1/6 + 1/3) = I R (2/6 + 1/6 + 2/6) = I R (5/6)$Since $V = I R_{eq}$,we have $I R_{eq} = I R (5/6)$.Therefore,the equivalent resistance is $R_{eq} = \frac{5}{6} R$.

If the resistance of each edge of a cube-shaped wire frame as shown in the figure below is $R$,then the resistance between points $1$ and $7$ is

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