JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Current Electricity

The four branches each supply $27$ V through $3\Omega$ (the third branch's $14$ V and $13$ V add to $27$ V), so in parallel they act as a single source of emf $27$ V and internal resistance $\frac{3}{4}\Omega$. This drives current through the $3\Omega +3\Omega =6\Omega$ path joining A and B, so $I=\frac{27}{6+3/4}=\frac{27}{27/4}=4$ A. The voltage across AB is $V=I\times 6=24$ V.