Electric Circuit Analysis/Kirchhoff's Current Law/Answers Exercise 6: 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!