PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 251

and

X

jC j j

CC

 Yj





11

10 3 5

1

6

6

 ()( )

The circuit of Figure 3.23 is then redrawn with the elements replaced by its equiva-

lent admittances and the current source by its phasor representation, as indicated

in Figure 3.24.

The two node equations are

For node E 1 , 5 ∠ 0 ° = (10 − j5)E 1 + j5E 2

For node E 2 , 0 = j5E 1 + (j + 3)E 2

The corresponding matrix equation is shown as follows:

()

()

10 5 5

53

50

0

1

2





jj 

jj

E

E











⋅













 ∠











10 Ω−^1

+

5 ∠ 0 °

j 6 Ω−^13 Ω− 1

−j 5 Ω−^1

E 1 E 2

FIGURE 3.24

Phasor network of R.3.58.

R 1 = 1/10 Ω

+

A –

5 cos (10 t)

C = 3/5 F R 2 = 1/3 Ω

L = 1/50 H

E 1 E 2

FIGURE 3.23

Network of R.3.58.