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(B) Let the nodes be labeled. The $4 \ \Omega$ resistor is connected between $A$ and the right junction. The $6 \ \Omega$ resistor is connected betwee

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$7$

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(B) Let the nodes be labeled. The $4 \ \Omega$ resistor is connected between $A$ and the right junction. The $6 \ \Omega$ resistor is connected between $B$ and the right junction.The two $12 \ \Omega$ resistors are connected in parallel between the right junction and the nodes $A$ and $B$ respectively.Let the right junction be $C$. The circuit consists of:$1$. $A$ $4 \ \Omega$ resistor between $A$ and $C$.$2$. $A$ $6 \ \Omega$ resistor between $B$ and $C$.$3$. $A$ $12 \ \Omega$ resistor between $A$ and $C$.$4$. $A$ $12 \ \Omega$ resistor between $B$ and $C$.Now,between $A$ and $C$,we have two resistors of $4 \ \Omega$ and $12 \ \Omega$ in parallel. Their equivalent resistance $R_{AC} = \frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3 \ \Omega$.Between $B$ and $C$,we have two resistors of $6 \ \Omega$ and $12 \ \Omega$ in parallel. Their equivalent resistance $R_{BC} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \ \Omega$.Finally,the equivalent resistance between $A$ and $B$ is the series combination of $R_{AC}$ and $R_{BC}$,which is $R_{AB} = 3 \ \Omega + 4 \ \Omega = 7 \ \Omega$.

In the given network,the equivalent resistance between $A$ and $B$ is .............. $\Omega$.

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