Electric Circuit Analysis/Kirchhoff's Current Law/Answers

From The Diagram

 From Node b we get: 
 * $$V_b = -V_1 = -15V $$

From Node d we get:
 * $$V_d = V_2 = -7V $$

It is clear that we must solve V_c, in order to complete Voltage definitions at all nodes. V_c will be found by applying KCL at Node c and solving resulting equations Follows: $$i_3 = i_1 + i_2 $$ $$ \frac{V_c}{R_3} = \frac{V_b - V_c}{R_1} + \frac{V_d - V_c}{R_2}$$

We can group like terms to get the following equation:

$$ V_c(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}) = \frac{V_b}{R_1} + \frac{V_d}{R_2}$$

Substitute values into previous equations you get:

$$V_c(\frac{1}{20\Omega} + \frac{1}{5\Omega} + \frac{1}{10\Omega}) = \frac{-15V}{20\Omega} + \frac{-7V}{5\Omega}$$ $$V_c(0.35) = -2.15 $$  thus   $$V_c = -6.14V  $$

Thus now we can calculate Current through $$R_3$$ as follows: $$\begin{matrix}\ I_{R3} &=& \frac{V_c}{R_3} \\ \ \\ \ &=& \frac{-6.14}{10} \\ \ \\ \ & = & -0.614A \end{matrix}$$.

Thus the effective current through $$R_3$$ is in opposite direction to $$i_3$$ Just as we expected!