CHAP. 4]

51

ANALYSIS METHODS

Similarly, I2 ¼

4.8

N2 1700 ¼ 3:17 A ¼ 536 R

I3 ¼

N3 5600 ¼ 10:45 A ¼ 536 R

Solve Problem 4.7 by the node voltage method. The circuit has been redrawn in Fig. 4-22, with two principal nodes numbered 1 and 2 and the third chosen as the reference node. By KCL, the net current out of node 1 must equal zero.

Fig. 4-22 V1 V1  25 V1  V2 þ ¼0 þ 2 10 5 Similarly, at node 2, V2  V1 V2 V2 þ 50 þ þ ¼0 10 4 2 Putting the two equations in matrix form,  2 32 3    1 1 1 1 5 V     6 2 þ 5 þ 10 76 1 7  10  6 76 7 ¼    4 4 5 5 1 1 1 1   þ þ  V2 25   10 10 4 2 The determinant of coeﬃcients and the numerator determinants are    0:80 0:10   ¼ 0:670  ¼  0:10 0:85      5 0:10   0:80  ¼ 1:75 N1 ¼  N2 ¼   25 0:85 0:10

 5  ¼ 19:5 25 

From these, V1 ¼

1:75 ¼ 2:61 V 0:670

V2 ¼

19:5 ¼ 29:1 V 0:670

In terms of these voltages, the currents in Fig. 4-21 are determined as follows: I1 ¼

4.9

V1 ¼ 1:31 A 2

I2 ¼

V1  V2 ¼ 3:17 A 10

I3 ¼

V2 þ 50 ¼ 10:45 A 2

For the network shown in Fig. 4-23, ﬁnd Vs which makes I0 ¼ 7:5 mA. The node voltage method will be used and the matrix form of the equations written by inspection.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An