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(D) To find the equivalent resistance between $A$ and $B$,we simplify the circuit from the right side.$1$. The last section consists of a $4\,\Omega$

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$8$

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(D) To find the equivalent resistance between $A$ and $B$,we simplify the circuit from the right side.$1$. The last section consists of a $4\,\Omega$ resistor in parallel with the series combination of the horizontal resistors. However,looking at the circuit,the rightmost vertical resistor is $4\,\Omega$. The horizontal resistors in the last loop are $1\,\Omega$ each. $2$. Let's simplify from right to left:- The rightmost loop has a $4\,\Omega$ resistor in parallel with the series combination of $(1+1+1+1)\,\Omega = 4\,\Omega$. - Equivalent resistance $R_1 = (4\,\Omega \parallel 4\,\Omega) = 2\,\Omega$.$3$. Now,this $2\,\Omega$ is in series with the next horizontal resistors $(1+1)\,\Omega$ on top and $(1+1)\,\Omega$ on bottom,making the total resistance in the next loop: $R_{next} = 2 + 2 + 2 = 6\,\Omega$.$4$. This $6\,\Omega$ is in parallel with the $8\,\Omega$ vertical resistor: $R_2 = (6\,\Omega \parallel 8\,\Omega) = \frac{6 	imes 8}{6 + 8} = \frac{48}{14} = \frac{24}{7}\,\Omega$.$5$. Adding the remaining horizontal resistors $(1+1)\,\Omega$ on top and $(1+1)\,\Omega$ on bottom: $R_{total} = \frac{24}{7} + 2 + 2 = \frac{24 + 28}{7} = \frac{52}{7}\,\Omega$.$6$. Finally,this is in parallel with the first $8\,\Omega$ resistor: $R_{CD} = (\frac{52}{7} \parallel 8) = \frac{\frac{52}{7} 	imes 8}{\frac{52}{7} + 8} = \frac{416}{52 + 56} = \frac{416}{108} \approx 3.85\,\Omega$.$7$. Adding the initial $2\,\Omega$ resistors in series: $R_{AB} = 2 + 3.85 + 2 = 7.85\,\Omega \approx 8\,\Omega$.

What is the effective resistance between $A$ and $B$ in the given figure (in $\Omega$)?

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