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(C) Let the four resistances be connected in a square $ABCD$ with vertices $A, B, C, D$. Each side has a resistance $R = 100 \ \Omega$.To find the eff

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$100$

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(C) Let the four resistances be connected in a square $ABCD$ with vertices $A, B, C, D$. Each side has a resistance $R = 100 \ \Omega$.To find the effective resistance across the diagonal points $A$ and $C$,we observe the circuit.The path $A-B-C$ consists of two resistors in series: $R_{ABC} = 100 \ \Omega + 100 \ \Omega = 200 \ \Omega$.The path $A-D-C$ also consists of two resistors in series: $R_{ADC} = 100 \ \Omega + 100 \ \Omega = 200 \ \Omega$.These two branches ($R_{ABC}$ and $R_{ADC}$) are connected in parallel between points $A$ and $C$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_{ABC}} + \frac{1}{R_{ADC}} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} = \frac{1}{100}$.Therefore,$R_{eq} = 100 \ \Omega$.

Four resistances of $100 \ \Omega$ each are connected in the form of a square. Then,the effective resistance along the diagonal points is ............. $\Omega$.

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