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(B) The circuit can be simplified step-by-step by identifying series and parallel combinations.$1$. The resistors of $2 \ \Omega$ and $2 \ \Omega$ in

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$\frac{8}{3} \ \Omega$

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(B) The circuit can be simplified step-by-step by identifying series and parallel combinations.$1$. The resistors of $2 \ \Omega$ and $2 \ \Omega$ in branch $ACD$ are in series. Their equivalent resistance is $R_1 = 2 + 2 = 4 \ \Omega$.$2$. This $R_1 = 4 \ \Omega$ is in parallel with the $4 \ \Omega$ resistor connected between $A$ and $D$. The equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $R_2 = 2 \ \Omega$.$3$. Now,$R_2 = 2 \ \Omega$ is in series with the $2 \ \Omega$ resistor in branch $DE$. Their equivalent resistance is $R_3 = 2 + 2 = 4 \ \Omega$.$4$. This $R_3 = 4 \ \Omega$ is in parallel with the $4 \ \Omega$ resistor connected between $A$ and $E$. The equivalent resistance $R_4$ is given by $\frac{1}{R_4} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $R_4 = 2 \ \Omega$.$5$. Finally,$R_4 = 2 \ \Omega$ is in series with the $2 \ \Omega$ resistor in branch $EB$. Their equivalent resistance is $R_5 = 2 + 2 = 4 \ \Omega$.$6$. This $R_5 = 4 \ \Omega$ is in parallel with the $8 \ \Omega$ resistor connected directly between $A$ and $B$. The total equivalent resistance $R_{AB}$ is $\frac{1}{R_{AB}} = \frac{1}{4} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8}$.$7$. Therefore,$R_{AB} = \frac{8}{3} \ \Omega$.

What is the equivalent resistance between $A$ and $B$ in the given circuit?

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