The net resistance between X and Y is ....... | vedclass

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(C) $1$. Analyze the upper branch: There are three $30 \ \Omega$ resistors in parallel. Their equivalent resistance $R_1 = \frac{30}{3} = 10 \ \Omega$

Vedclass · Accepted Answer

$15$

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(C) $1$. Analyze the upper branch: There are three $30 \ \Omega$ resistors in parallel. Their equivalent resistance $R_1 = \frac{30}{3} = 10 \ \Omega$.$2$. The $10 \ \Omega$ resistor on the right is short-circuited by the wire connected across it,so it is ignored.$3$. The upper branch now consists of a $10 \ \Omega$ resistor (from $R_1$) in series with a $20 \ \Omega$ resistor,giving $R_{upper} = 10 + 20 = 30 \ \Omega$.$4$. Analyze the lower branch: The $30 \ \Omega$ resistor is short-circuited by the wire connected across it,so it is ignored. The lower branch consists of a $20 \ \Omega$ resistor.$5$. The $10 \ \Omega$ resistor on the far left is in series with the combination of the upper and lower branches. However,looking at the circuit,the terminals $X$ and $Y$ are connected such that the total resistance between them is the parallel combination of the upper branch $(30 \ \Omega)$ and the lower branch ($20 \ \Omega$ in series with the $10 \ \Omega$ resistor on the left,total $30 \ \Omega$).$6$. Thus,$R_{eq} = \frac{30 	imes 30}{30 + 30} = 15 \ \Omega$.

The net resistance between $X$ and $Y$ is ............... $\Omega$.

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