Electric Circuit Analysis/Kirchhoff's Voltage Law/Answers
Exercise 5: Answers
Substitute the Above Result into (2)

Exercise 5: Answers
${\begin{matrix}\ I_{1}&=&{\frac {V_{2}\times R_{3}V_{1}\times (R_{2}+R_{3})}{(R_{3}^{2})(R_{2}+R_{3})(R_{3}+R_{1})}}\\\ \\\ &=&{\frac {(70)(15)(15)}{(100)(30)(15)}}\\\ \\\ &=&0.443A\end{matrix}}$
Substitute the Above Result into (2) ${\begin{matrix}\ I_{2}&=&{\frac {I_{1}*R_{3}V_{2}}{R_{2}+R_{3}}}&or&I_{2}&=&{\frac {I_{1}*(R_{3}+R_{1})V_{1}}{R_{3}}}\\\ \\\ &=&{\frac {(0.443A)(10\Omega )(7V)}{15\Omega }}&or&&=&{\frac {(0.443A)(30\Omega )(15V)}{10\Omega }}\\\ \\\ &=&0.171A\end{matrix}}$
${\begin{matrix}\ I_{R3}&=&(I_{1}I_{2})\\\ \\\ &=&0.443A(0.171A)\\\ \\\ &=&0.614A\end{matrix}}$
