Electric Circuit Analysis/Kirchhoff's Voltage Law/Answers

From KVL we get the following:

 $\begin{matrix} I_{1} & = & \frac{V_{2} \times R_{3} - V_{1} \times (R_{2} + R_{3})}{(R_{3}^{2}) - (R_{2} + R_{3}) (R_{3} + R_{1})} \\ = & \frac{(70) - (15) (15)}{(100) - (30) (15)} \\ = & 0.443 A \end{matrix}$

Substitute the Above Result into (2)

 $\begin{matrix} I_{2} & = & \frac{I_{1} * R_{3} - V_{2}}{R_{2} + R_{3}} & o r & I_{2} & = & \frac{I_{1} * (R_{3} + R_{1}) - V_{1}}{R_{3}} \\ = & \frac{(0.443 A) (10 Ω) - (7 V)}{15 Ω} & o r & = & \frac{(0.443 A) (30 Ω) - (15 V)}{10 Ω} \\ = & - 0.171 A \end{matrix}$

Thus now we can calculate Current through $R_{3}$ as follows:

 $\begin{matrix} I_{R 3} & = & (I_{1} - I_{2}) \\ = & 0.443 A - (- 0.171 A) \\ = & 0.614 A \end{matrix}$

Thus Current through $R_{3}$ is effectively flowing in the same direction as $I_{1}$ . See why you need to understand Passive sign Convention?

Template:Robelbox/close

Electric Circuit Analysis/Kirchhoff's Voltage Law/Answers

Navigation menu

Search