The effective resistance between A and B in t | vedclass

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(C) Let the resistance of the link $AB$ be $R_{AB} = 1 \Omega$. Let $R_{eq}$ be the effective resistance of the remaining part of the cube when the li

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$\frac{7}{5} \Omega$

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(C) Let the resistance of the link $AB$ be $R_{AB} = 1 \Omega$. Let $R_{eq}$ be the effective resistance of the remaining part of the cube when the link $AB$ is removed.When the link $AB$ is present,it is in parallel with the rest of the cube's network. The equivalent resistance is given by:$\frac{1}{R_{total}} = \frac{1}{R_{AB}} + \frac{1}{R_{eq}}$Given $R_{total} = \frac{7}{12} \Omega$ and $R_{AB} = 1 \Omega$:$\frac{12}{7} = \frac{1}{1} + \frac{1}{R_{eq}}$$\frac{1}{R_{eq}} = \frac{12}{7} - 1 = \frac{5}{7}$Therefore,$R_{eq} = \frac{7}{5} \Omega$.

The effective resistance between $A$ and $B$ in the figure is $\frac{7}{12} \Omega$ if each side of the cube has $1 \Omega$ resistance. The effective resistance between the same two points,when the link $A B$ is removed,is

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