JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Current Electricity

From the figure the three upper rows are identical: each has three $5$ V cells and three $3\Omega$ resistors in series, giving a branch emf of $15$ V and a branch resistance of $9\Omega$, and all three connect the same two rails $A$ and $B$ with their cells aiding. Three identical $15$ V, $9\Omega$ branches in parallel are equivalent to a single $15$ V source of internal resistance $\frac{9}{3}=3\Omega$. The bottom row, three $3\Omega$ resistors in series, is a $9\Omega$ path across the same terminals. The current through it is $I=\frac{15}{3+9}=1.25$ A, so the current between $A$ and $B$ is $1.25$ A.