Introduction to Electric Circuits

9. 7 The inverse transmission parameters 217

The receiving-end current is given by

I2 = 500 kVA/2 kV = 250 A

Because the receiving-end power factor is unity, I2 is in phase with V2 and so

12 = (250 + j0) A

From Equation (9.47)

V 1 -" V 2 "Jr- ZI:

Therefore

V1 = (2000 + j0) + (250 + j0)(0.22 + j0.36)

= 2000 + 55 + j90

= 2055 + j90

= V'(20552 + 902)/_tan -a (90/2055)

= 2057/_2.51 ~ V

Example 9.6

Determine the ABCD-parameters for the network shown in Fig. 9.9.

+o I1

Vl

--0

Z

1 I

I2 O+

2

O--

Figure 9.9

Solution

With the output port open circuited to make 12- 0, I~ = VIY and V2 = V1.

From Equation (9.36), A- V1/V2- 1. From Equation (9.38), C- I1/V2 =

I1/V 1 -- y.

With the output port short circuited to make V2 - 0, we see, using the current

division technique, that

I2-II[(1/Y)/{(1//Y) + Z}]

= I~[(1/Y)/{(1 + ZY)/Y}]

= 11/(1 + ZY)

From Equation (9.39), D - 11/12, so

D=I+ZY

From Equation (9.37), B = V1/I 2. Now

I~- V~/[{(1/Y)(Z)/((1/Y)+ Z)}]- VI[(1 + ZY)/Y]/(Z/Y) = VI(1 + ZY)/Z