Q 12-03-171JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
The current $I$ in the given circuit will be
Answer: (A) $10\ \text{A}$
Label the nodes: left $L$, right $R$, and the two junctions $M$, $N$ on either side of the $2\ \Omega$ resistor. Then $L$–$M = 4\ \Omega$, $L$–$N = 4\ \Omega$ (top bypass), $M$–$N = 2\ \Omega$, $N$–$R = 4\ \Omega$ and $M$–$R = 4\ \Omega$ (bottom bypass).
This is a balanced Wheatstone bridge ($\frac44 = \frac44$), so no current flows in the $2\ \Omega$ resistor:
$$R_{eq} = (4+4)\parallel(4+4) = 4\ \Omega,\qquad I = \frac{40}{4} = 10\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics