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(D) Let all resistors be $R$. The circuit consists of a triangle formed by $R_1, R_2, R_3$ connected to two external resistors $R_5$ and $R_4$ at term

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$8\,\Omega$

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(D) Let all resistors be $R$. The circuit consists of a triangle formed by $R_1, R_2, R_3$ connected to two external resistors $R_5$ and $R_4$ at terminals $A$ and $B$.First,consider the branch containing $R_1$ and $R_2$ in series. Their equivalent resistance is $R_s = R_1 + R_2 = R + R = 2R$.This combination is in parallel with $R_3$. Let the equivalent resistance of this part be $R_{eq1}$.$\frac{1}{R_{eq1}} = \frac{1}{2R} + \frac{1}{R} = \frac{1+2}{2R} = \frac{3}{2R} \implies R_{eq1} = \frac{2R}{3}$.Now,the total equivalent resistance across $A$ and $B$ is the series combination of $R_5$,$R_{eq1}$,and $R_4$.$R_{total} = R_5 + R_{eq1} + R_4 = R + \frac{2R}{3} + R = 2R + \frac{2R}{3} = \frac{8R}{3}$.Given $R = 3\,\Omega$,we have $R_{total} = \frac{8 	imes 3}{3} = 8\,\Omega$.

What is the equivalent resistance across $A$ and $B$ in the figure shown,if $R = 3\,\Omega$?

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