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: