20 Irwin, Engineering Circuit Analysis, 11e ISV
9
20 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 3: Nodal and Loop Analysis Techniques SOLUTION:
Transcript of 20 Irwin, Engineering Circuit Analysis, 11e ISV
20 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
67 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
68 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
101 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
The current through the 2 Ω resistor is i1
Mesh 1: 5i1
– 3i2
= 0
Mesh 2: –212 +8i 2–3i
1 = 0
Mesh 3: 8i3
– 5i2
+ 122 = 0
Solving, i 1= 20.52 A, i
2= 34.19 A and i
3 = 6.121 A
The current through the 5 Ω resistor is i3, or 6.121 A.
106 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
140 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
147 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
148 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
157 Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
SOLUTION:
![FAN7711 Ballast Control Integrated Circuit - Digi-Key Sheets/Fairchild PDFs/FAN7711.pdf · FAN7711 Ballast Control Integrated Circuit) 1 3 0 circuit [.] ...](https://static.fdocument.org/doc/165x107/5acfdb947f8b9a1d328d8e40/fan7711-ballast-control-integrated-circuit-digi-key-sheetsfairchild-pdfsfan7711pdffan7711.jpg)
