Electric Circuit Analysis/Kirchhoff's Current Law/Answers

< Electric Circuit Analysis < Kirchhoff's Current Law

Exercise 6: Answers

Model Answer

From The Diagram

From Node b we get:

V_b = -V_1 = -15V

From Node d we get:

V_d = V_2 = -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:

$i_3 = i_1 + i_2$

$\frac{V_c}{R_3} = \frac{V_b - V_c}{R_1} + \frac{V_d - V_c}{R_2}$

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:

 $V_c(\frac{1}{20\Omega} + \frac{1}{5\Omega} + \frac{1}{10\Omega}) = \frac{-15V}{20\Omega} + \frac{-7V}{5\Omega}$

 $V_c(0.35) = -2.15$    thus    $V_c = -6.14V$

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

Thus the effective current through $R_3$ is in opposite direction to $i_3$ Just as we expected!

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