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y 22
I2 2
y22 � �� E
The determining network for y12 appears in Fig. 26.41. Y1 is short
circuited; so IY2 � I1, and
IY2 � I1 ��E2Y2 The minus sign results because the defined direction of I 1 in Fig. 26.41 is
opposite to the actual flow direction due to the applied source E 2;that is,
y 12 �
E 1 � 0
I 1
� E2
E 1 � 0
(siemens, S) (26.35)
y22 � short-circuit, output-admittance parameter
The required network appears in Fig. 26.38.
1
E 1 = 0
1′
I 1
System
FIG. 26.38
y 22 determination.
EXAMPLE 26.9 Determine the admittance parameters for the p network
of Fig. 26.39.
Solution:
with
The network for y11 will appear as shown in Fig. 26.40,
Y1 � 0.2 mS �0° Y2 � 0.02 mS ��90° Y3 � 0.25 mS �90°
We use I1 � E1YT � E1(Y1 � Y2) with y 11 �
E 2 � 0
and y11 � Y1 � Y2 (26.36)
E 1
+
–
I 1
1
1′
Y 1
Y 2
I 1
� E1
Y 3
FIG. 26.40
Determining y 11.
2′
2
I 2
2
2′
+
–
E 2
Short circuited
E 2 = 0
ADMITTANCE (y) PARAMETERS ⏐⏐⏐ 1171
1 +
E 1
–
1′
I 1
B L
0.02 mS
G 0.2 mS
B C
FIG. 26.39
p network.
I 2
0.25 mS
2
+
E 2
–
2′