Electric Circuit Analysis/Nodal Analysis/Answers

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Exercise 7: Answers

Model Answer

KCL @ Node b:

 $i_2 = i_1 + i_6$

Thus by applying Ohms law to above equation we get.

 $\begin{matrix}\ \frac{V_c - V_b}{R_2} - \frac{V_b}{R_1} - \frac{V_b - V_d}{R_6} = 0 \end{matrix}$

Therefore

 $\begin{matrix}\ \ -V_b(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_6}) + V_c(\frac{1}{R_2}) + V_d(\frac{1}{R_6}) = 0\end{matrix}$    ...............   (1)

KCL @ Node c:

 $i_3 = i_2 + i_4$

Thus by applying Ohms law to above equation we get.

 $\begin{matrix}\ \frac{V_s - V_c}{R_3} - \frac{V_c - V_b}{R_2} - \frac{V_c - V_d}{R_4} = 0\end{matrix}$

Therefore

 $\begin{matrix}\ \ -V_b( \frac{1}{R_2}) - V_c((\frac{1}{R_2}+\frac{1}{R_3} + \frac{1}{R_4}) + V_d(\frac{1}{R_4}) = \frac{V_s}{R_3} \end{matrix}$    ...............   (2)

KCL @ Node d:

 $i_5 = i_4 + i_6$

Thus by applying Ohms law to above equation we get.

 $\begin{matrix}\ \frac{V_s - V_c}{R_3} - \frac{V_c - V_b}{R_2} - \frac{V_c - V_d}{R_4} \end{matrix}$

Therefore

 $\begin{matrix}\ \ -V_b(\frac{1}{R_3}) - V_c(\frac{1}{R_4}) + V_d(\frac{1}{R_3}+\frac{1}{R_4}-\frac{1}{R_5}) = 0 \end{matrix}$    ...............   (3)

$G_1 = \frac{1}{R_1}$ etc thus equations 1; 2 & 3 will be re-written as follows:

 $\begin{matrix}\ V_b(-G_1-G_2-G_6) + V_c(G_2) + V_d(G_6) &=& 0 ....(1) \\ \ \\ \ V_b(-G_2) + V_c(G_2+G_3+G_4) + V_d(-G_4) &=& (V_s \times G_3) ....(2) \\ \ \\ \ V_b(-G_3) + V_c(-G_4) + V_d(G_3+G_4+G_5) &=& 0 ....(3) \end{matrix}$

Now we can create a matrix with the above equations as follows:

 $\begin{bmatrix} (-G_1-G_2-G_6) & (G_2) & (G_6) \\ (-G_2) & (G_2+G_3+G_4) & (-G_4) \\ (-G_4) & (G_4) & (G_3+G_4+G_5)\end{bmatrix} . \begin{bmatrix} V_b\\ V_c \\ V_d \end{bmatrix} = \begin{bmatrix} 0 \\ V_s \times G_3 \\ 0 \end{bmatrix}$

The following matrix is the above with values substituted:

$A.\vec{X} = \vec{Y}$ → $\begin{bmatrix} -0.056 & 0.05 & 0.001 \\ -0.05 & 0.0503 & -0.0002 \\ -0.0001 & -0.0002 & 0.000367\end{bmatrix} . \begin{bmatrix} V_b\\ V_c \\ V_d \end{bmatrix} = \begin{bmatrix} 0 \\ 0.0009 \\ 0 \end{bmatrix}$

Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:

Solving determinants of:

Matrix A : General matrix A from KCL equations

Matrix A1 : Genral Matrix A with Column 1 substituted by $\vec Y$ .

Matrix A2 : Genral Matrix A with Column 2 substituted by $\vec Y$ .

Matrix A3 : Genral Matrix A with Column 3 substituted by $\vec Y$ .

As follows:

 $det A = -9.8 \times 10^{-8}$

 $det A1 = -1.7 \times 10^{-8}$

 $det A2 = -1.8 \times 10^{-8}$

 $det A3 = -1.5 \times 10^{-8}$

Now we can use the solved determinants to arrive at solutions for Node voltages $V_b; V_c and V_d$ as follows:

1. $V_b = \frac{det A1}{det A} = 0.17V$

2. $V_c = \frac{det A2}{det A} = 0.19V$

3. $V_d = \frac{det A3}{det A} = 0.15V$

Now we can apply Ohm's law to solve for the current through $R_3$ as follows:

 $I_{R_3} = \frac{V_s - V_c}{R_3} = \frac{9V - 0.19V}{10k\Omega} = 0.881mA$

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