Four wires of equal length and of resistances | vedclass

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(A) Let the four resistors be connected in a square $ABCD$. The resistance of each wire is $R = 10 \ \Omega$.When we find the equivalent resistance be

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$10$

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(A) Let the four resistors be connected in a square $ABCD$. The resistance of each wire is $R = 10 \ \Omega$.When we find the equivalent resistance between two opposite corners (say $D$ and $B$),the circuit splits into two parallel branches.Branch $1$ consists of two resistors in series: $DA$ and $AB$. The resistance of this branch is $R_1 = 10 \ \Omega + 10 \ \Omega = 20 \ \Omega$.Branch $2$ consists of two resistors in series: $DC$ and $CB$. The resistance of this branch is $R_2 = 10 \ \Omega + 10 \ \Omega = 20 \ \Omega$.Since these two branches are in parallel,the equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.Therefore,$R_{eq} = 10 \ \Omega$.

Four wires of equal length and of resistances $10 \ \Omega$ each are connected in the form of a square. The equivalent resistance between two opposite corners of the square is ............. $\Omega$. (in $\Omega$)

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