Electric Circuit Analysis/Kirchhoff's Current Law/Answers

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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:

$V_c(\frac{1}{20\mathrm{\Omega }}+\frac{1}{5\mathrm{\Omega }}+\frac{1}{10\mathrm{\Omega }})=\frac{-15V}{20\mathrm{\Omega }}+\frac{-7V}{5\mathrm{\Omega }}$

$V_c(0.35)=-2.15$   thus   $V_c=-6.14V$

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

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