Q 12-03-152JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
The equivalent resistance between $A$ and $B$ of the network shown in figure is
Answer: (D) $\dfrac83R$
Call the node after $2R$ $X$ and the node after $9R$ $Y$. Arms: $A$–$X$ $2R$, $A$–$Y$ $R$, $X$–$B$ $6R$, $Y$–$B$ $3R$, with $9R$ between $X$ and $Y$.
$\dfrac{2R}{6R}=\dfrac{R}{3R}$, so the bridge is balanced and $9R$ carries no current:
$$R_{AB}=(2R+6R)\parallel(R+3R)=\frac{8R\times4R}{12R}=\frac83R$$
Solution by Sreeraj P, M.Sc Physics