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GATE 2025 EC – Question 40

Networks, Signals and Systems · Two-Port Networks and Wye-Delta Transformation · 2 marks · Multiple choice

The $Z$-parameter matrix of a two port network relates the port voltages and port currents as follows:

$$\begin{bmatrix}V_1\\V_2\end{bmatrix}=Z\begin{bmatrix}I_1\\I_2\end{bmatrix}$$

The $Z$-parameter matrix (with each entry in Ohms) of the network shown below is ________.

Diagram for GATE 2025 EC question 40
  1. $\begin{bmatrix}\frac{10}{3}&\frac{2}{3}\\\frac{2}{3}&\frac{10}{3}\end{bmatrix}$
  2. $\begin{bmatrix}\frac{2}{3}&\frac{10}{3}\\\frac{10}{3}&\frac{2}{3}\end{bmatrix}$
  3. $\begin{bmatrix}10&2\\2&10\end{bmatrix}$
  4. $\begin{bmatrix}\frac{10}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{10}{3}\end{bmatrix}$

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Correct answer: (A) $\begin{bmatrix}\frac{10}{3}&\frac{2}{3}\\\frac{2}{3}&\frac{10}{3}\end{bmatrix}$

Explanation

With port 2 open ($I_2=0$): looking in from port 1, the first 2 Ω shunt is in parallel with (2 Ω series + 2 Ω shunt) $=4\ \Omega$, giving $2\parallel4=\frac43\ \Omega$; adding the input series 2 Ω, $Z_{11}=2+\frac43=\frac{10}{3}\ \Omega$. The voltage at the first shunt node is $\frac43I_1$; the open port voltage is the drop across the second shunt, half of that: $V_2=\frac23I_1$, so $Z_{21}=\frac23\ \Omega$. The network is symmetric and reciprocal, so $Z_{22}=\frac{10}{3}$ and $Z_{12}=\frac23$.