Kirchhoff's Voltage Law/Answers

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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 R3 as follows:

\begin{matrix}\ I_{R3} &=& (I_1 - I_2) \\ \ \\ \ &=& 0.443A - (-0.171A) \\ \ \\ \ & = & 0.614A \end{matrix}



Thus Current through R3 is effectively flowing in the same direction as I1. See why you need to understand Passive sign Convention?
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