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Solved Problems-11

For the network shown in the figure, determine the ABCD parameters.
Description: Description: 2973.png
The ABCD-parameter equations are,
Description: Description: 6294.png
For the network shown in the figure. we convert the delta consisting of the resistances of 2 Ω each into its equivalent star so that the circuit becomes as shown in Fig. (a) and Fig. (b).
Description: Description: 6303.png
Description: Description: 2992.png
To find the ABCD parameters, we consider two situations:
When V2 = 0, i.e., port-2 is short-circuited
As shown in Fig. (c), by KVL we get,
1.67I1 + 0.67(I1 + I2) = V1
or, 2.33I1 + 0.67I2 = V1
and, 0.67(I1 + I2) + 1.67I2 = 0
or, Description: Description: 6313.png
∴ D = Description: Description: 6319.png
Description: Description: 3011.png
Putting this value in the first equations, we get,
Description: Description: 6327.png
When I2 = 0, i. e. port-2 is open-circuited
Here, no current will flow through the right side 1.67Ω resistance. By KVL, we get,
V1 = (1.67 + 0.67)I1 = 2.33I1
and, V2 = 0.67I1
∴ Description: Description: 6333.png
∴ Description: Description: 6340.png
Description: Description: 3037.png
Therefore, the ABCD parameters of the network are
A = 3.5; B = 7.5Ω; C = 15; and D = 3.5

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