JEE Main · 2015 · 04 Apr 2015 · Offline · Current Electricity

Set the potential at $Q$ to zero and let the junction $P$ sit at potential $V$, then apply Kirchhoff's current law there: $\frac{V+6}{3}+\frac{V}{1}+\frac{V-9}{5}=0$. Clearing denominators and solving gives $V=-\frac{3}{23}\approx -0.13$V. The current through the $1\Omega$ resistor is $|V|/1\approx 0.13$A, and since $P$ sits below $Q$ in potential, that current runs from $Q$ to $P$.