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Y T � Y 2 � Y 3 and I 2 � E 2(Y 2 � Y 3)
Thus y 22 �
E 1 � 0
and y22 � Y2 � Y3 (26.39)
Substituting values, we have
I 2
� E2
Y1 � 0.2 mS �0°
Y2 � 0.02 mS ��90°
Y3 � 0.25 mS �90°
y11 � Y1 � Y2 � 0.2 mS � j 0.02 mS (L)
y12 � y21 ��Y2 ��(�j 0.02 mS)
� j 0.02 mS (C)
y22 � Y2 � Y3 ��j 0.02 mS � j 0.25 mS
� j 0.23 mS (C)
Note the similarities between the results for y 11 and y 22 for the p network
compared with z 11 and z 22 for the T network.
Two networks satisfying the terminal relationships of Eqs. (26.31a)
and (26.31b) are shown in Fig. 26.44. Note the use of parallel branches
1
+
E 1
–
1′
I 1
y 11
a
y 12 E 2
b
y 21 E 1
y 22
I 2
2
+
E 2
–
2′
1
+
E 1
–
1′
ADMITTANCE (y) PARAMETERS ⏐⏐⏐ 1173
y 11 + y 12
–y 12
y 22 + y 12
(a) (b)
FIG. 26.44
Two possible two-port, y-parameter equivalent networks.
since each term of Eqs. (26.31a) and (26.31b) has the units of current,
and the most direct route to the equivalent circuit is an application of
Kirchhoff’s current law in reverse. That is, find the network that satisfies
Kirchhoff’s current law relationship. For the impedance parameters,
each term had the units of volts, so Kirchhoff’s voltage law was
applied in reverse to determine the series combination of elements in
the equivalent circuit of Fig. 26.44(a).
Applying Kirchhoff’s current law to the network of Fig. 26.44(a), we
have
I 1
(y 22 – y 12 )I 1
I 2
2
+
E 2
–
2′