The figure below shows a part of an electric circuit. The potentials at points $$a$$, $$b$$ and $$c$$ are 30 V, 12 V and 2 V respectively. The current through the 20 $$\Omega$$ resistor will be,

Let the potential at the central junction be $$V_x$$.

$$I_{in} = I_{out1} + I_{out2}$$ (KCL)

Current from $$a$$ ($$I_{in}$$): $$\frac{30 - V_x}{10}$$

Current to $$b$$ ($$I_{out1}$$): $$\frac{V_x - 12}{20}$$

Current to $$c$$ ($$I_{out2}$$): $$\frac{V_x - 2}{30}$$

$$\frac{30 - V_x}{10} = \frac{V_x - 12}{20} + \frac{V_x - 2}{30}$$

$$6(30 - V_x) = 3(V_x - 12) + 2(V_x - 2)$$

$$180 - 6V_x = 3V_x - 36 + 2V_x - 4$$

$$V_x = 20\text{ V}$$

Current in $${20\Omega}$$ resistor:

$$I_{20\Omega} = \frac{V_x - V_b}{20}$$

$$I_{20\Omega} = \frac{20 - 12}{20} = \frac{8}{20}$$

$$I_{20\Omega} = 0.4\text{ A}$$