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ANALYSIS METHODS

[CHAP. 4

Fig. 4-20 32 3 2 3 I1 60 7 7 13 7 54 I2 5 ¼ 4 60 5 60 I3 7 19 2 3 19 7 7 6 7 R ¼ 4 7 13 7 5 ¼ 2880 7 7 19

2

19 4 7 7 Thus,

Notice that in Problem 4.2, too, R ¼ 2880, although the elements in the determinant were different. All valid sets of meshes or loops yield the same numerical value for R . The three numerator determinants are    60 7 7   N2 ¼ 8642 7  ¼ 4320 N3 ¼ 4320 N1 ¼  60 13  60 7 19  Consequently, I1 ¼

N1 4320 ¼ 1:5 A ¼ R 2880

I2 ¼

N2 ¼ 3A R

I3 ¼

N3 ¼ 1:5 A R

The current supplied by the 60-V source is the sum of the three loop currents, I1 þ I2 þ I3 ¼ 6 A.

4.7

Write the mesh current matrix equation for the network of Fig. 4-21 by inspection, and solve for the currents.

Fig. 4-21 2

7 4 5 0

5 19 4

3 32 3 2 25 0 I1 4 54 I2 5 ¼ 4 25 5 I3 50 6

Solving,      25 5 0  0   7 5     I1 ¼  25 19 4    5 19 4  ¼ ð700Þ  536 ¼ 1:31 A  50 4 6 6   0 4

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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