Electric Circuit Analysis/Nodal Analysis/Answers

Template:Robelbox Template:Image 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:

 $[\begin{matrix} (- G_{1} - G_{2} - G_{6}) & (G_{2}) & (G_{6}) \\ (- G_{2}) & (G_{2} + G_{3} + G_{4}) & (- G_{4}) \\ (- G_{4}) & (G_{4}) & (G_{3} + G_{4} + G_{5}) \end{matrix}] . [\begin{matrix} V_{b} \\ V_{c} \\ V_{d} \end{matrix}] = [\begin{matrix} 0 \\ V_{s} \times G_{3} \\ 0 \end{matrix}]$

The following matrix is the above with values substituted:

$A . \vec{X} = \vec{Y}$ → $[\begin{matrix} - 0.056 & 0.05 & 0.001 \\ - 0.05 & 0.0503 & - 0.0002 \\ - 0.0001 & - 0.0002 & 0.000367 \end{matrix}] . [\begin{matrix} V_{b} \\ V_{c} \\ V_{d} \end{matrix}] = [\begin{matrix} 0 \\ 0.0009 \\ 0 \end{matrix}]$

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:

 $d e t A = - 9.8 \times 1 0^{- 8}$

 $d e t A 1 = - 1.7 \times 1 0^{- 8}$

 $d e t A 2 = - 1.8 \times 1 0^{- 8}$

 $d e t A 3 = - 1.5 \times 1 0^{- 8}$

Now we can use the solved determinants to arrive at solutions for Node voltages $V_{b}; V_{c} a n d V_{d}$ as follows:

1. $V_{b} = \frac{d e t A 1}{d e t A} = 0.17 V$

2. $V_{c} = \frac{d e t A 2}{d e t A} = 0.19 V$

3. $V_{d} = \frac{d e t A 3}{d e t A} = 0.15 V$

Now we can apply Ohm's law to solve for the current through $R_{3}$ as follows:

 $I_{R_{3}} = \frac{V_{s} - V_{c}}{R_{3}} = \frac{9 V - 0.19 V}{10 k Ω} = 0.881 m A$

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Electric Circuit Analysis/Nodal Analysis/Answers

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