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(A) Let the nodes be $X$,$A$,$B$,and $Y$. Based on the circuit diagram:$1$. The $10 \, \Omega$ resistor is connected between $X$ and $A$.$2$. The $10

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$10 \, \Omega$

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(A) Let the nodes be $X$,$A$,$B$,and $Y$. Based on the circuit diagram:$1$. The $10 \, \Omega$ resistor is connected between $X$ and $A$.$2$. The $10 \, \Omega$ resistor is connected between $X$ and $B$.$3$. The $20 \, \Omega$ resistor is connected between $A$ and $B$.$4$. The $10 \, \Omega$ resistor is connected between $A$ and $Y$.$5$. The $10 \, \Omega$ resistor is connected between $B$ and $Y$.This is a Wheatstone bridge configuration. Since the ratio of resistors in the arms is $\frac{10}{10} = \frac{10}{10}$,the bridge is balanced.Therefore,no current flows through the $20 \, \Omega$ resistor connected between $A$ and $B$.The circuit simplifies to two parallel branches,each containing two $10 \, \Omega$ resistors in series:Branch $1$ $(X-A-Y)$: $10 \, \Omega + 10 \, \Omega = 20 \, \Omega$Branch $2$ $(X-B-Y)$: $10 \, \Omega + 10 \, \Omega = 20 \, \Omega$Now,these two $20 \, \Omega$ branches are in parallel:$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$$R_{eq} = 10 \, \Omega$

What is the equivalent resistance between $X$ and $Y$ in the given circuit?

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