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.