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(A) To find the equivalent resistance between points $X$ and $Y$,we first simplify the circuit by identifying parallel and series combinations.$1$. Th

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(A) To find the equivalent resistance between points $X$ and $Y$,we first simplify the circuit by identifying parallel and series combinations.$1$. The circuit can be redrawn by identifying nodes. Let the node between the middle resistors be $Z$.$2$. The resistors connected in parallel between $X$ and $Z$ have a resistance of $R_1 = \frac{2 	imes 2}{2 + 2} = 1 \Omega$.$3$. Similarly,the resistors connected in parallel between $Z$ and $Y$ have a resistance of $R_2 = \frac{2 	imes 2}{2 + 2} = 1 \Omega$.$4$. Now,$R_1$ and $R_2$ are in series,so their equivalent resistance is $R_3 = R_1 + R_2 = 1 + 1 = 2 \Omega$.$5$. Finally,this $R_3$ is in parallel with the top resistor of $2 \Omega$ connected directly between $X$ and $Y$. Thus,the total equivalent resistance $R_{eq}$ is:$R_{eq} = \frac{R_3 	imes 2}{R_3 + 2} = \frac{2 	imes 2}{2 + 2} = \frac{4}{4} = 1 \Omega$.

In the network shown in the figure,the equivalent resistance between points $X$ and $Y$ will be . . . . . . $\Omega$. The value of each resistor is $2 \Omega$.

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