Kirchhoff's Current Law/Answers

From Wikiversity

Jump to: navigation, search
Artículo bueno.svg
Exercise 6: Answers


EE-102-L06-Fig4.jpg

Model Answer

From The Diagram

From Node b we get:

Vb = − V1 = − 15V

From Node d we get:

Vd = V2 = − 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:

i3 = i1 + i2

 \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}


Vc(0.35) =  − 2.15   thus   Vc =  − 6.14V

Thus now we can calculate Current through R3 as follows:
\begin{matrix}\ I_{R3} &=& \frac{V_c}{R_3} \\ \ \\ \ &=& \frac{-6.14}{10} \\ \ \\ \ & = & -0.614A \end{matrix}.


Thus the effective current through R3 is in opposite direction to i3 Just as we expected!