Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Question

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Answer 1

Question

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

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Answer 2

Question

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

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Answer 3

Question

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

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Answer 4

Question

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

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Answer 5

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Answer 6

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)

i2= sprintf('%.7f A',val2)

i3= sprintf('%.7f A',val3)

i4= sprintf('%.3f A',val4)

i5= sprintf('%.7f A',val5)

The output in command window:

i1 = '0.0016196 A'

i2 = '-0.0010202 A'

i3 = '-0.0024612 A'

i4 = '-0.003 A'

i5 = '-0.0024230 A'

Convert obtained value of currents in milli amperes (mA).

i1=0.0016196 A=1.6196×10−3 A {∵1×10−3 A=1 mA}=1.6196 mA

i2=−0.0010202 A=−1.0202×10−3 A=−1.0202 mA

i3=−0.0024612 A=−2.4612×10−3 A=−2.4612 mA

i4=−0.003 A=−3×10−3 A=−3 mA

i5=−0.002423 A=−2.423×10−3 A=−2.423 mA

Conclusion:

Therefore, the value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Answer 7

Chapter 3, Problem 53P

To determine

Find the mesh currents in the circuit of Figure 3.98 using MATLAB.

Answer 8

Expert Solution & Answer

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196 mA, −1.0202 mA, −2.4612 mA, −3 mA, and −2.423 mA respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

Refer to Figure 3.98 in the textbook for mesh analysis.

Calculation:

From Figure 3.98, write the expression for current i4.

i4=−3 mA (1)

Apply Kirchhoff’s voltage law to loop 1 with current i1 in Figure 3.98.

(1 kΩ)(i1−i3)+(3 kΩ)(i1−i2)−12=01000i1−1000i3+3000i1−3000i2=12 {∵1 kΩ=1000 Ω}4000i1−3000i2−1000i3=12

4i1−3i2−i3=0.012 (2)

Apply Kirchhoff’s voltage law to loop 2 with current i2 in Figure 3.98.

(4 kΩ)(i2−i4)+(3 kΩ)(i2−i1)=04000i2−4000i4+3000i2−3000i1=0−3000i1+7000i2−4000i4=0

−3i1+7i2−4i4=0 (3)

Substitute equation (1) in (3).

−3i1+7i2−4(−3 mA)=0−3i1+7i2−4(−3×10−3)=0 {∵1 mA=1×10−3 A}−3i1+7i2+0.012=0

−3i1+7i2=−0.012 (4)

Apply Kirchhoff’s voltage law to loop 3 with current i3 in Figure 3.98.

(1 kΩ)(i3−i1)+(6 kΩ)(i3−i5)+(8 kΩ)(i3−i4)=01000i3−1000i1+6000i3−6000i5+8000i3−8000i4=0−1000i1+15000i3−8000i4−6000i5=0

−i1+15i3−8i4−6i5=0 (5)

Substitute equation (1) to (5).

−i1+15i3−8(−3 mA)−6i5=0−i1+15i3+0.024−6i5=0

−i1+15i3−6i5=−0.024 (6)

Apply Kirchhoff’s voltage law to loop 5 with current i5 in Figure 3.98.

(2 kΩ)i5+(8 kΩ)(i5−i4)+(6 kΩ)(i5−i3)=02000i5+8000i5−8000i4+6000i5−6000i3=0−6000i3−8000i4+16000i5=0

−3i3−4i4+8i5=0 (7)

Substitute equation (1) to (7).

−3i3−4(−3 mA)+8i5=0−3i3+0.012+8i5=0

−3i3+8i5=−0.012 (8)

MATLAB code:

Write the MATLAB code to solve the equations (2), (4), (6), and (8) as follows in MATLAB code editor, and save it as “node353”, and then run the code. The result will shows in main command window.

syms i1 i2 i3 i5

eq1 = 4*i1 -3*i2 -1*i3 +0*i5 == 0.012;

eq2 = -3*i1 +7*i2 +0*i3 +0*i5 == -0.012;

eq3 = -1*i1 +0*i2 +15*i3 -6*i5 == -0.024;

eq4 = 0*i1 +0*i2 -3*i3 +8*i5 == -0.012;

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

val1 = sol.i1;

val2 = sol.i2;

val3 = sol.i3;

val4 = -0.003;

val5 = sol.i5;

i1= sprintf('%.7f A',val1)