Consider the two-port network as shown in the Figure. Which of the following options provides the correct set of values of A, B, C and D parameters?

The correct option is:

$A=2,B=3\times {10}^3\Omega ,C={10}^{-3}{\Omega }^{-1},D=2$

Step-by-Step Explanation:

1. Analysis of the Circuit Diagram:
From the given image, we observe a T-network representing the two-port network with Port 1 (P1) on the left and Port 2 (P2) on the right.
The values of the three resistors are:

• Left series branch: $Z_1=1\text{k}\Omega =1000\Omega$
• Right series branch: $Z_3=1\text{k}\Omega =1000\Omega$
• Shunt (middle) branch: $Z_2=1\text{k}\Omega =1000\Omega$

2. Two-Port Network Definitions:
The transmission (ABCD) parameters of a two-port network relate the input variables ($V_1$, $I_1$) to the output variables ($V_2$, $I_2$) according to the following equations:

$V_1=A{V}_2-B{I}_2$

$I_1=C{V}_2-D{I}_2$

Here, $I_1$ is the current entering Port 1, and $I_2$ is the current entering Port 2. (Alternatively, if $I_2$ is defined as leaving Port 2, the minus signs become plus signs).

3. Calculating parameters A and C (Open-Circuit at Port 2):
We open-circuit Port 2, which means the current leaving Port 2 is zero, so $I_2=0$.
Since Port 2 is open, no current flows through the right-arm resistor $Z_3$. The voltage at the output port, $V_2$, is equal to the voltage across the shunt resistor $Z_2$.
Applying a voltage divider at Port 1:

$V_2=V_1\times \frac{Z_2}{Z_1+Z_2}=V_1\times \frac{1000}{1000+1000}=\frac{1}{2}{V}_1$

Therefore, the parameter A is:

$A={\frac{V_1}{V_2}}_{I_2=0}=2$

Next, the input current $I_1$ under this open-circuit condition flows entirely through $Z_1$ and $Z_2$:

$I_1=\frac{V_2}{Z_2}$

Therefore, the parameter C is:

$C={\frac{I_1}{V_2}}_{I_2=0}=\frac{1}{Z_2}=\frac{1}{1000}={10}^{-3}{\Omega }^{-1}$

4. Calculating parameters B and D (Short-Circuit at Port 2):
We short-circuit Port 2, meaning $V_2=0$.
Let $I_2^{\text{'}}=-I_2$ be the current flowing out of Port 2 under this condition. Since Port 2 is shorted, the resistors $Z_2$ and $Z_3$ are connected in parallel with respect to the node between them. Using current division:

$I_2^{\text{'}}=I_1\times \frac{Z_2}{Z_2+Z_3}=I_1\times \frac{1000}{1000+1000}=\frac{1}{2}{I}_1$

Thus, the parameter D is:

$D={\frac{I_1}{-I_2}}_{V_2=0}=\frac{I_1}{I_2^{\text{'}}}=2$

To find B, we write the input voltage $V_1$ under this short-circuit condition. The total equivalent impedance seen from Port 1 is:

$R_{\text{eq}}=Z_1+(Z_2\parallel Z_3)=1000+\frac{1000\times 1000}{1000+1000}=1500\Omega$

Hence, $V_1=I_1\times 1500$.
Since $I_1=2I_2^{\text{'}}$, we substitute to find $V_1$ in terms of $I_2^{\text{'}}$:

$V_1=\left(2I_2^{\text{'}}\right)\times 1500=3000I_2^{\text{'}}$

Therefore, the parameter B is:

$B={\frac{V_1}{-I_2}}_{V_2=0}=\frac{V_1}{I_2^{\text{'}}}=3000\Omega =3\times {10}^3\Omega$

5. Summary of Parameter Values:
Collecting all the calculated transmission parameters, we obtain:

• $A=2$
• $B=3\times {10}^3\Omega$
• $C={10}^{-3}{\Omega }^{-1}$
• $D=2$