Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents. Figure 3.110 For Prob. 3.66.

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Answer 1

Textbook Question

Chapter 3, Problem 66P

Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents.

Chapter 3, Problem 66P, Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh

Figure 3.110

For Prob. 3.66.

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

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Answer 2

Textbook Question

Chapter 3, Problem 66P

Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents.

Chapter 3, Problem 66P, Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh

Figure 3.110

For Prob. 3.66.

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

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Answer 3

Textbook Question

Chapter 3, Problem 66P

Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents.

Chapter 3, Problem 66P, Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh

Figure 3.110

For Prob. 3.66.

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

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Answer 4

Textbook Question

Chapter 3, Problem 66P

Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents.

Chapter 3, Problem 66P, Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh

Figure 3.110

For Prob. 3.66.

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Answer 8

Expert Solution & Answer

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;

i1= sprintf('%.4f A',val1)

i2= sprintf('%.4f A',val2)

i3= sprintf('%.4f A',val3)

i4= sprintf('%.4f A',val4)

i5= sprintf('%.4f A',val5)

The output in command window:

i1 = '-0.2779 A'

i2 = '-1.0488 A'

i3 = '1.4682 A'

i4 = '-0.4761 A'

i5 = '-2.2332 A'

Conclusion:

Therefore, the matrix form of mesh current equations is,

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32].

And the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Write the mesh current equations, and find the mesh currents in the circuit of Figure 3.110 using mesh analysis and MATLAB.

Answer 12

Answer to Problem 66P

The matrix form of mesh current equations is [30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32], and the value of currents i1, i2, i3, i4, and i5 in the given circuit are −0.2779 A, −1.0488 A, 1.4682 A, −0.4761 A, and −2.2332 A respectively.

Answer 13

Explanation of Solution

Given data:

Refer Figure 3.110 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.110.

10i1+4(i1−i2)+24+2(i1−i4)+6(i1−i3)−12+8i1=010i1+4i1−4i2+2i1−2i4+6i1−6i3+8i1=12−24

30i1−4i2−6i3−2i4=−12 (1)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.110.

10i2+8i2+40+6(i2−i5)+2(i2−i4)−24+4(i2−i1)=010i2+8i2+6i2−6i5+2i2−2i4+4i2−4i1=24−40

−4i1+30i2−2i4−6i5=−16 (2)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.110.

−30+8i3+6(i3−i1)+4(i3−i4)=0−30+8i3+6i3−6i1+4i3−4i4=0

−6i1+18i3−4i4=30 (3)

Apply Kirchhoff’s voltage law to loop 4 with current i4 in Figure 3.110.

2(i4−i1)+2(i4−i2)+4(i4−i5)+4(i4−i3)=02i4−2i1+2i4−2i2+4i4−4i5+4i4−4i3=0

−2i1−2i2−4i3+12i4−4i5=0 (4)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.110.

6(i5−i2)+8i5+32+4(i5−i4)=06i5−6i2+8i5+32+4i5−4i4=0

−6i2−4i4+18i5=−32 (5)

Represent the equations (1), (2), (3), (4), and (5) in matrix form.

[30−4−6−20−4300−2−6−6018−40−2−2−412−40−60−418][i1i2i3i4i5]=[−12−16300−32]

MATLAB code:

Write the MATLAB code to solve the equations (1), (2), (3), (4), and (5) as follows in MATLAB code editor, and save it as “node366”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i4 i5

eq1 = 30*i1 -4*i2 -6*i3 -2*i4 +0*i5 == -12;

eq2 = -4*i1 +30*i2 +0*i3 -2*i4 -6*i5 == -16;

eq3 = -6*i1 +0*i2 +18*i3 -4*i4 +0*i5 == 30;

eq4 = -2*i1 -2*i2 -4*i3 +12*i4 -4*i5 == 0;

eq5 = 0*i1 -6*i2 +0*i3 -4*i4 +18*i5 == -32;

sol = solve([eq1, eq2, eq3, eq4, eq5], [i1, i2, i3, i4, i5]);

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = sol.i4;

val5 = sol.i5;