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48

ANALYSIS METHODS

[CHAP. 4

Fig. 4-17 KVL and KCL give: I2 ð12Þ ¼ I3 ð6Þ I2 ð12Þ ¼ I4 ð12Þ

ð10Þ ð11Þ

60 ¼ I1 ð7Þ þ I2 ð12Þ I1 ¼ I2 þ I3 þ I4

ð12Þ ð13Þ

I1 ¼ I2 þ 2I2 þ I2 ¼ 4I2

ð14Þ

Substituting (10) and (11) in (13),

Now (14) is substituted in (12): 60 ¼ I1 ð7Þ þ 14 I1 ð12Þ ¼ 10I1

4.2

or

I1 ¼ 6 A

Solve Problem 4.1 by the mesh current method.

Fig. 4-18 Applying KVL to each mesh (see Fig. 4-18) results in 60 ¼ 7I1 þ 12ðI1  I2 Þ 0 ¼ 12ðI2  I1 Þ þ 6ðI2  I3 Þ 0 ¼ 6ðI3  I2 Þ þ 12I3 Rearranging terms and putting the equations in matrix form, ¼ 60 19I1  12I2 12I1 þ 18I2  6I3 ¼ 0  6I2 þ 18I3 ¼ 0

2 or

19 12 4 12 18 0 6

32 3 2 3 0 60 I1 6 54 I2 5 ¼ 4 0 5 18 I3 0

Using Cramer’s rule to find I1 ,     60 12 0   19   I1 ¼  0 18 6    12  0 6 18   0

 12 0  18 6  ¼ 17 280  2880 ¼ 6 A 6 18 

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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