Electric Circuit Analysis/Kirchhoff's Voltage Law/Answers

Exercise 5: Answers From KVL we get the following: ${\begin{matrix}\ I_{1}&=&{\frac {V_{2}\times R_{3}-V_{1}\times (R_{2}+R_{3})}{(R_{3}^{2})-(R_{2}+R_{3})(R_{3}+R_{1})}}\\\ \\\ &=&{\frac {(70)-(15)(15)}{(100)-(30)(15)}}\\\ \\\ &=&0.443A\end{matrix}}$ Substitute the Above Result into (2) ${\begin{matrix}\ I_{2}&=&{\frac {I_{1}*R_{3}-V_{2}}{R_{2}+R_{3}}}&or&I_{2}&=&{\frac {I_{1}*(R_{3}+R_{1})-V_{1}}{R_{3}}}\\\ \\\ &=&{\frac {(0.443A)(10\Omega )-(7V)}{15\Omega }}&or&&=&{\frac {(0.443A)(30\Omega )-(15V)}{10\Omega }}\\\ \\\ &=&-0.171A\end{matrix}}$ Thus now we can calculate Current through $R_{3}$ as follows: ${\begin{matrix}\ I_{R3}&=&(I_{1}-I_{2})\\\ \\\ &=&0.443A-(-0.171A)\\\ \\\ &=&0.614A\end{matrix}}$ Thus Current through $R_{3}$ is effectively flowing in the same direction as $I_{1}$. See why you need to understand Passive sign Convention?