Introduction to Electric Circuits

216 Two-port networks

To find B and D we short circuit the output port to make V2 = 0. Then
V1 = IzZ and I~ = I2. From Equation (9.37), B = V~/I2 = Z. From Equation
(9.39), D = I~/I2 = 1.
We could also find the ABCD-parameters by using Kirchhoff's laws as
follows. Applying KVL to the network and taking the clockwise direction to be
positive, we see that V1 - IzZ - V2 = 0. Rearranging we get

V 1 --" g 2 Jr- lI 2 (9.47)

Comparing Equations (9.33) and (9.47) we see that A = 1 and that B = Z.

Applying KCL to the circuit we see that 11 = 12, which may be written as

11 = OV2 + 12 (9.48)

Comparing Equations (9.34) and (9.48) we see that C = 0 and that D = 1.

In matrix form

V1

Note that A = D. This is always the case for symmetrical two-port networks,

that is networks for which the input and output ports are interchangeable. A

two-port network which is not symmetrical is shown in Example 9.6.

'Short' power transmission lines

A power transmission line will have conductor resistance and inductance

distributed along the length of the line and capacitance between conductors

also distributed along the length of the line. A 'short' line (less than about

80 km in length) is one for which, for the purposes of analysis, the capacitance

can be neglected and the resistance and inductance can be considered to be

concentrated at the centre of the line without introducing great errors. The line

may then be represented by the series impedance network of Fig. 9.8.

Example 9.5

A single-phase transmission line has an impedance Z = (0.22 + j0.36)11. (1)

determine the ABCD-parameters of this line, and (2) calculate the sending-end

voltage required to produce 500 kVA at a voltage of 2 kV when operating at

unity power factor.

Solution

The network is as shown in Fig. 9.8. From the matrix Equation (9.49) we see

that A = D = 1; B = Z = (0.22 + j0.36) 11; C = 0. Taking the receiving-end

voltage as the reference,

V2 = (2000 + jO) V