Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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On this question, I am supposed to find Iphi and V1. I Applied KCL at Node 1, and made an auxillary equation for the supernode? Should I have left all the terms in fractions for my matrix-
Where did I go wrong?
![Example 10: Determine \( v_1 \) and \( i_{\phi} \) in the circuit shown in Figure 19.
[Figure showing a circuit diagram]
Figure 19: Circuit for Example 10.
The circuit diagram consists of:
- A 60 V voltage source.
- Resistors labeled as 2 \( \Omega \) and 3 \( \Omega \) on the upper path.
- A dependent current source labeled as \( 6i_{\phi} \).
- Parallel paths with resistors labeled as 24 \( \Omega \) and another 3 \( \Omega \).
- Nodes labeled \( V_1 \) and \( V_2 \), with \( v_1 \) indicated across the 24 \( \Omega \) resistor.
**Auxiliary Equation:**
\[ V_2 - 6i_{\phi} = 60 \]
**Given:**
We know what \( v_3 \) is, so we don’t need to solve for it.
**KCL at \( V_1 \):**
\[ \frac{V_1 - 60}{2} + \frac{V_1}{24} + \frac{V_1 - V_2}{3} = 0 \]
- LCM is 24, leading to:
\[ 12V_1 - 720 + V_1 + 8V_1 - 8V_2 = 0 \]
\[ 21V_1 - 8V_2 = 720 \] *(Eq. 1)*
**Auxiliary Equation:**
\[ V_2 - 6\left( \frac{V_2 - V_1}{2} \right) = V_3 \]
\[ V_2 - \frac{6V_2 - 6V_1}{3} = 60 \]
\[ 3V_2 - 6V_2 + 6V_1 = 180 \]
\[ -6V_1 - 3V_2 = 180 \] *(Eq. 2)*
Conclusion:
- Solve Equations 1 and 2 to find \( v_1 \) and \( i_{\phi} \).
- Note \( V_3 = 60 \).](https://content.bartleby.com/qna-images/question/5362cbea-9427-4b2b-99cb-57a208edd853/c9116170-e085-4a6b-b638-b31b16bc79de/dqh3899_processed.jpeg)
Transcribed Image Text:Example 10: Determine \( v_1 \) and \( i_{\phi} \) in the circuit shown in Figure 19.
[Figure showing a circuit diagram]
Figure 19: Circuit for Example 10.
The circuit diagram consists of:
- A 60 V voltage source.
- Resistors labeled as 2 \( \Omega \) and 3 \( \Omega \) on the upper path.
- A dependent current source labeled as \( 6i_{\phi} \).
- Parallel paths with resistors labeled as 24 \( \Omega \) and another 3 \( \Omega \).
- Nodes labeled \( V_1 \) and \( V_2 \), with \( v_1 \) indicated across the 24 \( \Omega \) resistor.
**Auxiliary Equation:**
\[ V_2 - 6i_{\phi} = 60 \]
**Given:**
We know what \( v_3 \) is, so we don’t need to solve for it.
**KCL at \( V_1 \):**
\[ \frac{V_1 - 60}{2} + \frac{V_1}{24} + \frac{V_1 - V_2}{3} = 0 \]
- LCM is 24, leading to:
\[ 12V_1 - 720 + V_1 + 8V_1 - 8V_2 = 0 \]
\[ 21V_1 - 8V_2 = 720 \] *(Eq. 1)*
**Auxiliary Equation:**
\[ V_2 - 6\left( \frac{V_2 - V_1}{2} \right) = V_3 \]
\[ V_2 - \frac{6V_2 - 6V_1}{3} = 60 \]
\[ 3V_2 - 6V_2 + 6V_1 = 180 \]
\[ -6V_1 - 3V_2 = 180 \] *(Eq. 2)*
Conclusion:
- Solve Equations 1 and 2 to find \( v_1 \) and \( i_{\phi} \).
- Note \( V_3 = 60 \).
![This image contains a mathematical calculation involving matrices and determinants commonly found in linear algebra, particularly in solving systems of linear equations using Cramer's rule.
1. **Matrix Equation:**
\[
\begin{bmatrix}
21 & -8 \\
-6 & -3
\end{bmatrix}
\begin{bmatrix}
V_1 \\
V_2
\end{bmatrix}
=
\begin{bmatrix}
720 \\
180
\end{bmatrix}
\]
This represents a system of linear equations with unknowns \( V_1 \) and \( V_2 \).
2. **Determinant Calculation:**
The determinant \(\Delta\) of the coefficient matrix is computed as:
\[
\Delta = 21 \cdot (-3) - (-6) \cdot (-8) = -111
\]
3. **Cramer's Rule:**
- **Determinant \(\Delta_1\):**
Substitute the first column of the coefficient matrix with the constants:
\[
\Delta_1 =
\begin{vmatrix}
720 & -8 \\
180 & -3
\end{vmatrix}
=
720 \cdot (-3) - 180 \cdot (-8) = -720
\]
- **Determinant \(\Delta_2\):**
Substitute the second column of the coefficient matrix with the constants:
\[
\Delta_2 =
\begin{vmatrix}
21 & 720 \\
-6 & 180
\end{vmatrix}
=
21 \cdot 180 - (-6) \cdot 720 = 8100
\]
4. **Solving for Unknowns:**
Using Cramer's rule:
\[
V_1 = \frac{\Delta_1}{\Delta} = \frac{-720}{-111} = \frac{240}{37} \quad \text{(Note: should get 480)}
\]
\[
V_2 = \frac{\Delta_2}{\Delta} = \frac{8100}{-111} = -\frac{](https://content.bartleby.com/qna-images/question/5362cbea-9427-4b2b-99cb-57a208edd853/c9116170-e085-4a6b-b638-b31b16bc79de/4p2en7u_processed.jpeg)
Transcribed Image Text:This image contains a mathematical calculation involving matrices and determinants commonly found in linear algebra, particularly in solving systems of linear equations using Cramer's rule.
1. **Matrix Equation:**
\[
\begin{bmatrix}
21 & -8 \\
-6 & -3
\end{bmatrix}
\begin{bmatrix}
V_1 \\
V_2
\end{bmatrix}
=
\begin{bmatrix}
720 \\
180
\end{bmatrix}
\]
This represents a system of linear equations with unknowns \( V_1 \) and \( V_2 \).
2. **Determinant Calculation:**
The determinant \(\Delta\) of the coefficient matrix is computed as:
\[
\Delta = 21 \cdot (-3) - (-6) \cdot (-8) = -111
\]
3. **Cramer's Rule:**
- **Determinant \(\Delta_1\):**
Substitute the first column of the coefficient matrix with the constants:
\[
\Delta_1 =
\begin{vmatrix}
720 & -8 \\
180 & -3
\end{vmatrix}
=
720 \cdot (-3) - 180 \cdot (-8) = -720
\]
- **Determinant \(\Delta_2\):**
Substitute the second column of the coefficient matrix with the constants:
\[
\Delta_2 =
\begin{vmatrix}
21 & 720 \\
-6 & 180
\end{vmatrix}
=
21 \cdot 180 - (-6) \cdot 720 = 8100
\]
4. **Solving for Unknowns:**
Using Cramer's rule:
\[
V_1 = \frac{\Delta_1}{\Delta} = \frac{-720}{-111} = \frac{240}{37} \quad \text{(Note: should get 480)}
\]
\[
V_2 = \frac{\Delta_2}{\Delta} = \frac{8100}{-111} = -\frac{
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