JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Current Electricity

Take the lower rail as the zero of potential and let $V$ be the potential of the junction at the top of the $6\Omega$ resistor. Both sources feed that junction through their series resistances, and the whole of the current leaves through the $6\Omega$ branch:

Multiplying through by $60$ gives $3(140-V)+12(90-V)=10V$, so $1500=25V$ and $V=60$ V. The current in the $6\Omega$ resistor is $\frac{60}{6}=10$ A.