Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Question
Chapter 3, Problem 53P
To determine

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

Expert Solution & Answer
Check Mark

Answer to Problem 53P

The value of currents i1, i2, i3, i4, and i5 are 1.6196mA, 1.0202mA, 2.4612mA, 3mA, and 2.423mA 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=3mA (1)

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

(1)(i1i3)+(3)(i1i2)12=01000i11000i3+3000i13000i2=12{1=1000Ω}4000i13000i21000i3=12

4i13i2i3=0.012 (2)

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

(4)(i2i4)+(3)(i2i1)=04000i24000i4+3000i23000i1=03000i1+7000i24000i4=0

3i1+7i24i4=0 (3)

Substitute equation (1) in (3).

3i1+7i24(3mA)=03i1+7i24(3×103)=0{1mA=1×103A}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)(i3i1)+(6)(i3i5)+(8)(i3i4)=01000i31000i1+6000i36000i5+8000i38000i4=01000i1+15000i38000i46000i5=0

i1+15i38i46i5=0 (5)

Substitute equation (1) to (5).

i1+15i38(3mA)6i5=0i1+15i3+0.0246i5=0

i1+15i36i5=0.024 (6)

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

(2)i5+(8)(i5i4)+(6)(i5i3)=02000i5+8000i58000i4+6000i56000i3=06000i38000i4+16000i5=0

3i34i4+8i5=0 (7)

Substitute equation (1) to (7).

3i34(3mA)+8i5=03i3+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=0.0016196A=1.6196×103A{1×103A=1mA}=1.6196mA

i2=0.0010202A=1.0202×103A=1.0202mA

i3=0.0024612A=2.4612×103A=2.4612mA

i4=0.003A=3×103A=3mA

i5=0.002423A=2.423×103A=2.423mA

Conclusion:

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

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Chapter 3 Solutions

Fundamentals of Electric Circuits

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