Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

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Answer 1

Textbook Question

Chapter 4, Problem 14E

Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44.

Chapter 4, Problem 14E, Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

FIGURE 4.44

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

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SA [(a) 5 V (b) 5 V] 13. Find the voltage V in the network shown in Fig. 2.44 (a) if R is 10 2 and (b) 20 2 14. In the network of Fig. 2.44 (b), calculate the voltage between points a and b i.e. Vab [30 V] (Elect. Engg. I, Bombay Univ.) 4A 78A 4 h 10A ww 3A (a) ΤΑ 6A DC Network Theorems SA Is 1A 77 12A www 12 6A 8A Fig. 2.44 (b) [Hint: In the above two cases, the two closed loops are independent and no current passes between them].

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Answer 2

Textbook Question

Chapter 4, Problem 14E

Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44.

Chapter 4, Problem 14E, Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

FIGURE 4.44

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

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Answer 3

Textbook Question

Chapter 4, Problem 14E

Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44.

Chapter 4, Problem 14E, Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

FIGURE 4.44

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

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Answer 4

Textbook Question

Chapter 4, Problem 14E

Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44.

Chapter 4, Problem 14E, Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

FIGURE 4.44

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Answer 8

Expert Solution & Answer

To determine

Find the numerical value of each nodal voltage.

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the numerical value of each nodal voltage.

Answer 12

Answer to Problem 14E

The nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Answer 13

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node v1.

v12+v1−v36=26v1+2(v1−v3)12=26v1+2(v1−v3)=248v1−2v3=24

4v1−v3=12 (1)

Apply nodal analysis at node v2.

−2=v25+v2−v32+v2−v4102v2+5(v2−v3)+v2−v4=−20

8v2−5v3−v4=−20 (2)

Apply nodal analysis at node v3.

1=v3−v45+v3−v16+v3−v221=6(v3−v4)+5(v3−v1)+15(v3−v2)30

−5v1−15v2+26v3−6v4=30 (3)

Apply nodal analysis at node v4.

v4−v210+v4−v35+v45=0v4−v2+2(v4−v3)+2v4=0

−v2−2v3+5v4=0 (4)

Solve the equations by Cramer’s rule.

[12−20300]=[40−1008−5−1−5−1526−60−1−25][v1v2v3v4]

Find Δ.

Δ=|40−1008−5−1−5−1526−60−1−25|=1737

Find Δv1.

Δv1=|120−10−208−5−130−1526−60−1−25|=5346

The value of v1 is,

v1=Δv1Δ=53461737=3.078 V

Find Δv2.

Δv2=|412−100−20−5−1−53026−600−25|=−4080

The value of v2 is,

v2=Δv2Δ=−40801737=−2.349 V

Find Δv3.

Δv3=|4012008−20−1−5−1530−60−105|=540

The value of v3 is,

v3=Δv3Δ=5401737=0.3109 V

Find Δv4.

Δv4=|40−11208−5−20−5−1526300−1−20|=−600

The value of v4 is,

v4=Δv4Δ=−6001737=−0.3454 V

Apply nodal analysis at node v5.

v51+v5−v74=−24v5+v5−v74=−2

5v5−v7=−8 (5)

Apply nodal analysis at node v6.

v6−v72+v64+v6−v84=22(v6−v7)4+v64+v6−v84=22v6−2v7+v6+v6−v84=2

4v6−2v7−v8=8 (6)

Apply nodal analysis at node v7.

v7−v810+v7−v54+v7−v62=62(v7−v8)20+5(v7−v5)20+10(v7−v6)20=62v7−2v8+5v7−5v5+10v7−10v620=6

−5v5−10v6+17v7−2v8=120 (7)

Apply nodal analysis at node v8.

v8−v64+v81+v8−v710=010(v8−v6)40+40v840+4(v8−v7)40=010v8−10v6+40v8+4v8−4v740=0

−10v6−4v7+54v8=0 (8)

Solve the equations by Cramer’s rule.

[50−1004−2−1−5−1017−20−10−454][v5v6v7v8]=[−881200]

Find Δ.

Δ=|50−1004−2−1−5−1017−20−10−454|=10,520

Find Δv5.

Δv5=|−80−1088−5−1120−1526−60−1−25|=10,720

The value of v5 is,

v5=Δv5Δ=10,72010,520=1.02 V

Find Δv6.

Δv6=|5−8−1008−2−1−512017−200−454|=96,960

The value of v6 is,

v6=Δv6Δ=96,96010,520=9.217 V

Find Δv7.

Δv7=|50−80048−1−5−10120−20−10054|=137,760

The value of v7 is,

v7=Δv7Δ=137,76010,520=13.095 V

Find Δv8.

Δv8=|50−1−804−28−5−10171200−10−40|=28,160

The value of v8 is,

v8=Δv8Δ=28,16010,520=2.6768 V

Conclusion:

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	v1	3.078 V
2	v2	–2.349 V
3	v3	0.3109 V
4	v4	–0.3454 V
5	v5	1.02 V
6	v6	9.217 V
7	v7	13.095 V
8	v8	2.6768 V

Answer 14

Ch. 4.1 - For the circuit of Fig. 4.3, determine the nodal...Ch. 4.1 - For the circuit of Fig. 4.5, compute the voltage...Ch. 4.1 - For the circuit of Fig. 4.8, determine the nodal...Ch. 4.2 - For the circuit of Fig. 4.11, compute the voltage...Ch. 4.3 - Determine i1 and i2 in the circuit in Fig. 4.19....Ch. 4.3 - Determine i1 and i2 in the circuit of Fig 4.21....Ch. 4.3 - Determine i1 in the circuit of Fig. 4.24 if the...Ch. 4.4 - Determine the current i1 in the circuit of Fig....Ch. 4.4 - Determine v3 in the circuit of Fig. 4.28. FIGURE...Ch. 4 - Solve the following systems of equations: (a) 2v2 ...

Determine a numerical value for each nodal voltage in the circuit of Fig. 4.44. FIGURE 4.44

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Answer to Problem 14E

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