JEE Main 2026 Apr 5 Shift 1, Physics Q13: Kirchhoff’s Laws

Each of the three upper rows reduces to a single branch with EMF $3\times 5=15$ V and internal resistance $3\times 3=9\Omega$. The bottom row with no cell is a $9\Omega$ resistor.

The three EMF branches in parallel give an equivalent EMF of $15$ V (all identical) and equivalent internal resistance $\frac{9}{3}=3\Omega$.

This combination is connected through the external $9\Omega$ branch between A and B. Total resistance in the loop: $3+9=12\Omega$.

Current from A to B: $I_{AB}=\frac{15}{12}=1.25$ A.