In the circuit shown, the current in the 1 $$\Omega$$ resistor is:

Connect the lower wire of the $$1\ \Omega$$ resistor to the ground ($$Q$$ end) and apply KCL at node $$P$$.

$$\frac{V + 6}{3} + \frac{V}{1} + \frac{V - 9}{5} = 0$$

$$\Rightarrow V \left[ \frac{1}{3} + 1 + \frac{1}{5} \right] = \frac{9}{5} - \frac{6}{3}$$

$$\Rightarrow V \left[ \frac{5 + 15 + 3}{15} \right] = \frac{9 - 10}{5}$$

$$\Rightarrow V \left[ \frac{23}{15} \right] = -\frac{1}{5}$$

$$\Rightarrow V = -\frac{1}{5} \times \frac{15}{23} = \frac{-3}{23} = -0.13\text{ V}$$

Thus, the magnitude of current in the $$1\ \Omega$$ resistor is $$I = \frac{|V|}{R} = \frac{0.13}{1} = 0.13\text{ A}$$

Since the potential at $$P$$ is negative relative to $$Q$$, the current flows from $$Q$$ to $$P$$.