Determine v 1 and v 2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

Question

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Answer 1

Textbook Question

Chapter 3, Problem 56P

Determine v₁ and v₂ in the circuit of Fig. 3.101.

Chapter 3, Problem 56P, Determine v1 and v2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

Figure 3.101

For Prob. 3.56.

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

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Answer 2

Textbook Question

Chapter 3, Problem 56P

Determine v₁ and v₂ in the circuit of Fig. 3.101.

Chapter 3, Problem 56P, Determine v1 and v2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

Figure 3.101

For Prob. 3.56.

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

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Answer 3

Textbook Question

Chapter 3, Problem 56P

Determine v₁ and v₂ in the circuit of Fig. 3.101.

Chapter 3, Problem 56P, Determine v1 and v2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

Figure 3.101

For Prob. 3.56.

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

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Answer 4

Textbook Question

Chapter 3, Problem 56P

Determine v₁ and v₂ in the circuit of Fig. 3.101.

Chapter 3, Problem 56P, Determine v1 and v2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

Figure 3.101

For Prob. 3.56.

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Answer 8

Expert Solution & Answer

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the voltages v1 and v2 in the circuit of Figure 3.101.

Answer 12

Answer to Problem 56P

The value of voltages v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Answer 13

Explanation of Solution

Given data:

Refer Figure 3.101 in the textbook for mesh analysis.

Calculation:

Redraw the given circuit as shown in Figure 1 with the mesh currents.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 56P

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 1.

20i1+10(i1−i2)+100+10(i1−i3)=020i1+10i1−10i2+10i1−10i3=−10040i1−10i2−10i3=−100

4i1−i2−i3=−10 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 1.

10i2+10(i2−i3)−100+10(i2−i1)=010i2+10i2−10i3−100+10i2−10i1=0−10i1+30i2−10i3=100

−i1+3i2−i3=10 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 1.

20i3+10(i3−i1)+10(i3−i2)=020i3+10i3−10i1+10i3−10i2=0−10i1−10i2+40i3=0

−i1−i2+4i3=0 (3)

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

(4−1−1−13−1−1−14)(i1i2i3)=(−10100)

Obtain the value of the determinants as follows.

Δ=|4−1−1−13−1−1−14|=4(12−1)+1(−4−1)−1(1+3)=44−5−4=35

Δ2=|4−10−1−110−1−104|=4(40−0)+10(−4−1)−1(0+10)=160−50−10=100

Δ3=|4−1−10−1310−1−10|=4(0+10)+1(0+10)−10(1+3)=40+10−40=10

Write the expression for the mesh currents using Cramer’s rule.

i2=Δ2Δ (4)

i3=Δ3Δ (5)

Substitute 100 for Δ2 and 35 for Δ in equation (4) to find i2.

i2=10035=2.857 A

Substitute 10 for Δ3 and 35 for Δ in equation (5) to find i3.

i3=1035=0.2857 A

With reference to Figure 1, write the expression for the voltages v1 and v2.

v1=(10 Ω)i2 (6)

v2=−(20 Ω)i3 (7)

Substitute 2.857 A for i2 in equation (6) to find v1.

v1=(10 Ω)(2.857 A)=28.57 V

Substitute 0.2857 A for i3 in equation (7) to find v2.

v2=−(20 Ω)(0.2857 A)=−5.714 V

Conclusion:

Therefore, the values of voltage v1 and v2 in the circuit are 28.57 V and −5.714 V respectively.

Answer 14

Ch. 3.2 - Figure 3.4 For Practice Prob. 3.1. Obtain the node...Ch. 3.2 - Figure 3.6 For Practice Prob. 3.2. Find the...Ch. 3.3 - Figure 3.11 For Practice Prob. 3.3. Find v and i...Ch. 3.3 - Figure 3.14 For Practice Prob. 3.4. Find v1, v2,...Ch. 3.4 - Practice Problem 3.5 Figure 3.19 For Practice...Ch. 3.4 - Practice Problem 3.6 Figure 3.21 For Practice...Ch. 3.5 - Practice Problem 3.7 Figure 3.25 For Practice...Ch. 3.6 - By inspection, obtain the node-voltage equations...Ch. 3.6 - By inspection, obtain the mesh-current equations...Ch. 3.8 - For the circuit in Fig. 3.33, use PSpice to find...

Determine v 1 and v 2 in the circuit of Fig. 3.101. Figure 3.101 For Prob. 3.56.

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