Twelve equal resistors,each of resistance R,a | vedclass

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(C) Let a total current $6I$ enter the cube at one corner. Due to symmetry,the current splits into three equal parts of $2I$ at each of the three edge

Vedclass · Accepted Answer

$5R/6$

Vedclass · Answer

(C) Let a total current $6I$ enter the cube at one corner. Due to symmetry,the current splits into three equal parts of $2I$ at each of the three edges connected to the entry point.At the next set of nodes,each current $2I$ splits into two equal parts of $I$ along the edges.At the next set of nodes,the currents recombine to form $2I$ in each of the three edges connected to the exit corner.Thus,the total current leaving the exit corner is $2I + 2I + 2I = 6I$.Applying Kirchhoff's Voltage Law along a path from the entry corner to the exit corner (e.g.,through three edges): $V = (2I)R + (I)R + (2I)R = 5IR$.The equivalent resistance $R_{eq}$ is given by $R_{eq} = V / I_{total} = 5IR / 6I = 5R/6$.

Twelve equal resistors,each of resistance $R$,are connected to form a skeleton cube. The equivalent resistance between two diagonally opposite corners is:

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