Each resistance in the given cubical network | vedclass

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(A) Let the total current entering the cubical network at point $A$ be $6I$. Due to the symmetry of the cube, the current splits equally into three br

Vedclass · Accepted Answer

$\frac{5}{6} \Omega$

Vedclass · Answer

(A) Let the total current entering the cubical network at point $A$ be $6I$. Due to the symmetry of the cube, the current splits equally into three branches at $A$, each carrying $2I$. At the next set of nodes, these currents split again. Applying Kirchhoff's voltage law to a path from $A$ to $B$ (e.g., $A ightarrow 	ext{node} ightarrow 	ext{node} ightarrow B$), the potential drop $V$ across the network is the sum of potential drops across the resistors in any path. For a path consisting of three edges, the potential drop is $V = (2I 	imes R) + (I 	imes R) + (2I 	imes R) = 5IR$. Given $R = 1 \Omega$, we have $V = 5I$. The equivalent resistance $R_{eq}$ is given by $R_{eq} = \frac{V}{I_{total}} = \frac{5I}{6I} = \frac{5}{6} \Omega$.

Each resistance in the given cubical network has a resistance of $1 \Omega$. The equivalent resistance between points $A$ and $B$ is:

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