The equivalent resistance between A and B is | vedclass

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(B) $1$. Identify the nodes in the circuit. Let the node at $A$ be $1$. The node after the first resistor is $2$. The node after the second resistor i

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$8$

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(B) $1$. Identify the nodes in the circuit. Let the node at $A$ be $1$. The node after the first resistor is $2$. The node after the second resistor is $3$. The node after the third resistor is $4$. The node after the fourth resistor is $5$. The node at $B$ is the end.$2$. Observe that the resistor between nodes $4$ and $5$ is short-circuited by a wire,so it can be ignored.$3$. Redraw the circuit: The resistors between $1-2$,$1-3$,$2-3$,$2-4$,and $3-4$ form a Wheatstone bridge structure.$4$. Specifically,nodes $2$ and $3$ are connected to $1$ and $4$ via $4 \ \Omega$ resistors. The resistor between $2$ and $3$ is $4 \ \Omega$. This is a balanced Wheatstone bridge.$5$. The equivalent resistance of the bridge part (between $1$ and $4$) is $4 \ \Omega$.$6$. Finally,this is in series with the last $4 \ \Omega$ resistor connected to $B$.$7$. Total equivalent resistance $R_{eq} = 4 \ \Omega + 4 \ \Omega = 8 \ \Omega$.

The equivalent resistance between $A$ and $B$ is ............ $\Omega$.

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