A circuit has four resistors and two voltage sources. The resistors are R1= 0.001, R₂ = 3.33, R3 = 4, and R4 = 0.012 ohms; the voltage sources are E₁ = 12 and E₂ = 10 volts; and the current are 1₁, 12, and i3 amps. By using Kirchohoff's laws, we have the linear system 0.013i₁ +3.331₂ 0.013i₁ 1₁ 1₂ = 22 + 413 = 12 13 = 0 (a) Use Gaussian elimination without pivoting and three-digit rounding to solve the system. (b) Use Gaussian elimination with partial pivoting and three-digit rounding to solve the system. (c) Use Gaussian elimination with complete pivoting and three-digit rounding to solve the system.
A circuit has four resistors and two voltage sources. The resistors are R1= 0.001, R₂ = 3.33, R3 = 4, and R4 = 0.012 ohms; the voltage sources are E₁ = 12 and E₂ = 10 volts; and the current are 1₁, 12, and i3 amps. By using Kirchohoff's laws, we have the linear system 0.013i₁ +3.331₂ 0.013i₁ 1₁ 1₂ = 22 + 413 = 12 13 = 0 (a) Use Gaussian elimination without pivoting and three-digit rounding to solve the system. (b) Use Gaussian elimination with partial pivoting and three-digit rounding to solve the system. (c) Use Gaussian elimination with complete pivoting and three-digit rounding to solve the system.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Circuit Analysis Using Gaussian Elimination**
**Introduction:**
A circuit consists of four resistors and two voltage sources. The resistors have the following values:
- \( R_1 = 0.001 \, \text{ohms} \)
- \( R_2 = 3.33 \, \text{ohms} \)
- \( R_3 = 4 \, \text{ohms} \)
- \( R_4 = 0.012 \, \text{ohms} \)
The voltage sources are given by:
- \( E_1 = 12 \, \text{volts} \)
- \( E_2 = 10 \, \text{volts} \)
The circuit also includes currents \( i_1, i_2, \) and \( i_3 \) in amps.
**Equations:**
Using Kirchhoff’s laws, the system can be represented by the following linear equations:
\[
0.013i_1 + 3.33i_2 = 22
\]
\[
0.013i_1 + 4i_3 = 12
\]
\[
i_1 - i_2 - i_3 = 0
\]
**Problem-Solving Approach:**
The task is to solve the linear system using Gaussian elimination. The steps are to be done under different conditions:
(a) Use Gaussian elimination without pivoting and three-digit rounding to solve the system.
(b) Use Gaussian elimination with partial pivoting and three-digit rounding to solve the system.
(c) Use Gaussian elimination with complete pivoting and three-digit rounding to solve the system.
**Conclusion:**
Each method for Gaussian elimination provides different levels of accuracy and stability, especially concerning rounding errors. Understanding and applying these methods can contribute to insightful analyses and solutions in electrical circuit problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2f2c4bd-bf4c-4a3b-a0a2-6333c3306a45%2F95df7b66-bd82-41bc-bc62-23dc21f4887d%2F9ygeoao_processed.png&w=3840&q=75)
Transcribed Image Text:**Circuit Analysis Using Gaussian Elimination**
**Introduction:**
A circuit consists of four resistors and two voltage sources. The resistors have the following values:
- \( R_1 = 0.001 \, \text{ohms} \)
- \( R_2 = 3.33 \, \text{ohms} \)
- \( R_3 = 4 \, \text{ohms} \)
- \( R_4 = 0.012 \, \text{ohms} \)
The voltage sources are given by:
- \( E_1 = 12 \, \text{volts} \)
- \( E_2 = 10 \, \text{volts} \)
The circuit also includes currents \( i_1, i_2, \) and \( i_3 \) in amps.
**Equations:**
Using Kirchhoff’s laws, the system can be represented by the following linear equations:
\[
0.013i_1 + 3.33i_2 = 22
\]
\[
0.013i_1 + 4i_3 = 12
\]
\[
i_1 - i_2 - i_3 = 0
\]
**Problem-Solving Approach:**
The task is to solve the linear system using Gaussian elimination. The steps are to be done under different conditions:
(a) Use Gaussian elimination without pivoting and three-digit rounding to solve the system.
(b) Use Gaussian elimination with partial pivoting and three-digit rounding to solve the system.
(c) Use Gaussian elimination with complete pivoting and three-digit rounding to solve the system.
**Conclusion:**
Each method for Gaussian elimination provides different levels of accuracy and stability, especially concerning rounding errors. Understanding and applying these methods can contribute to insightful analyses and solutions in electrical circuit problems.
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