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.

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.