For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection. Figure 3.113 For Prob. 3.69.

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Answer 1

Textbook Question

Chapter 3, Problem 69P

For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection.

Chapter 3, Problem 69P, For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection. Figure 3.113

Figure 3.113

For Prob. 3.69.

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

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Answer 2

Textbook Question

Chapter 3, Problem 69P

For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection.

Chapter 3, Problem 69P, For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection. Figure 3.113

Figure 3.113

For Prob. 3.69.

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

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Answer 3

Textbook Question

Chapter 3, Problem 69P

For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection.

Chapter 3, Problem 69P, For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection. Figure 3.113

Figure 3.113

For Prob. 3.69.

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

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Answer 4

Textbook Question

Chapter 3, Problem 69P

For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection.

Chapter 3, Problem 69P, For the circuit shown in Fig. 3.113, write the node-voltage equations by inspection. Figure 3.113

Figure 3.113

For Prob. 3.69.

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Answer 8

Expert Solution & Answer

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50

Simplify the equation as follows.

v2−v110+v2−v35+v210=50v210−v110+v25−v35+v210=50−v110+(110+15+110)v2−v35=50

−0.1v1+0.4v2−0.2v3=50 (2)

Apply Kirchhoff’s current law at node v3 in Figure 3.113.

v3−v120 kΩ+v3−v25 kΩ−40 mA+50 mA=0v3−v120 kΩ+v3−v25 kΩ=40 mA−50 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v3−v120+v3−v25=(−10×10−3)(103)v3−v120+v3−v25=−10

Simplify the equation as follows.

v320−v120+v35−v25=−10−v120−v25+(120+15)v3=−10

−0.05v1−0.2v2+0.25v3=−10 (3)

Represent the equations (1), (2), and (3) in matrix form.

[0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10]

Conclusion:

Thus, the matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Write the node-voltage equations for the circuit in Figure 3.113.

Answer 12

Answer to Problem 69P

The matrix form representation of node-voltage equations for the given circuit is [0.35−0.1−0.05−0.10.4−0.2−0.05−0.20.25][v1v2v3]=[10050−10].

Answer 13

Explanation of Solution

Given data:

Refer to Figure 3.113 in the textbook for the nodal analysis.

Calculation:

Apply Kirchhoff’s current law at node v1 in Figure 3.113.

v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ−100 mA=0v1−v320 kΩ+v1−v210 kΩ+v1−05 kΩ=100 mA (∵1 kΩ=1×103 Ω1 mA=1×10−3 A)v1−v320+v1−v210+v1−05=(100×10−3)(103)v1−v320+v1−v210+v1−05=100

Simplify the equation as follows.

v1−v320+v1−v210+v1−05=100v120−v320+v110−v210+v15=100(120+110+15)v1−v210−v320=100

0.35v1−0.1v2−0.05v3=100 (1)

Apply Kirchhoff’s current law at node v2 in Figure 3.113.

v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ−50 mA=0v2−v110 kΩ+v2−v35 kΩ+v2−010 kΩ=50 mA (∵1 kΩ=1×103 Ω, and 1 mA=1×10−3 A)v2−v110+v2−v35+v2−010=(50×10−3)(103)v2−v110+v2−v35+v2−010=50