GATE 2025 EC Q.40: solution and answer

GATE 2025 · ECQ.40 · 2 marks · MCQ

The $Z$ -parameter matrix of a two port network relates the port voltages and port currents as follows:
$[V_{1} V_{2}] = Z [I_{1} I_{2}]$
The $Z$ -parameter matrix (with each entry in Ohms) of the network shown below is ________.

GATE 2025 EC question 40: A ladder two-port network with a common bottom (ground) rail. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$[10/3 2/3 2/3 10/3]$
(B)
$[2/3 10/3 10/3 2/3]$
(C)
$[102210]$
(D)
$[10/3 1/3 1/3 10/3]$

Answer and solution

ANSWER

Option (A): $Z = [10/3 2/3 2/3 10/3] Ω$ .

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Definition of the Z-parameters. With port 2 open ( $I_{2} = 0$ ): $z_{11} = V_{1} / I_{1}$ and $z_{21} = V_{2} / I_{1}$ . With port 1 open ( $I_{1} = 0$ ): $z_{22} = V_{2} / I_{2}$ and $z_{12} = V_{1} / I_{2}$ . Call the node after the first series resistor A and the node after the second series resistor B.
$z_{11}$ (port 2 open). With $I_{2} = 0$ no current flows in the right-hand series $2 Ω$ , so the branch leaving node A to the right is just the middle series $2 Ω$ plus the shunt $2 Ω$ at B, i.e. $4 Ω$ . Seen from port 1: $z_{11} = 2 + (2 ∥ 4) = 2 + \frac{2 \cdot 4}{2 + 4} = 2 + \frac{4}{3} = \frac{10}{3} Ω$
$z_{21}$ (port 2 open), current divider at node A. $I_{1}$ splits between the shunt $2 Ω$ at A and the $4 Ω$ path. The current in the $4 Ω$ path (through the shunt at B) is $I_{A B} = I_{1} \cdot \frac{2}{2 + 4} = \frac{I _{1}}{3}$
Because $I_{2} = 0$ , there is no drop across the right-hand series resistor, so $V_{2}$ equals the voltage across the shunt at B: $V_{2} = V_{B} = 2 \cdot \frac{I _{1}}{3} = \frac{2}{3} I_{1} \Rightarrow z_{21} = \frac{2}{3} Ω$ (Check by nodal analysis: at B, $(V_{A} - V_{B}) /2 = V_{B} /2$ gives $V_{A} = 2 V_{B}$ ; at A, $I_{1} = V_{A} /2 + (V_{A} - V_{B}) /2 = \frac{3}{2} V_{B}$ , so $V_{B} = \frac{2}{3} I_{1}$ and $V_{1} = 2 I_{1} + V_{A} = \frac{10}{3} I_{1}$ .)
Reciprocity and symmetry. The network contains only resistors, so it is reciprocal: $z_{12} = z_{21} = \frac{2}{3} Ω$ . It is also a mirror image of itself about its centre (series 2, shunt 2, series 2, shunt 2, series 2), so it is symmetric: $z_{22} = z_{11} = \frac{10}{3} Ω$ .
Result. $Z = [10/3 2/3 2/3 10/3] Ω$ This is option (A).

KEY IDEA

Open-circuit one port at a time. With a port open, its series resistor carries no current, so it drops out of the input impedance and adds no drop to the output voltage. That makes $z_{11}$ a series-parallel reduction and $z_{21}$ a single current division. Reciprocity (the network has only resistors) and mirror symmetry then give $z_{12} = z_{21}$ and $z_{22} = z_{11}$ with no extra work.

WATCH OUT FOR

Each wrong option matches a specific slip. Adding all five resistors in series, as if the shunts were series elements, gives 10 and 2 (option C). Writing the divided current $I_{1} /3$ as $z_{21}$ and forgetting to multiply by the $2 Ω$ shunt at B gives 1/3 (option D). Swapping the driving-point and transfer entries gives option B: $z_{11}$ must be the larger one, since $V_{1}$ includes the drop across the first series resistor. Another trap is counting the open-circuited right-hand series resistor in $z_{21}$ . With $I_{2} = 0$ it has no drop across it.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2025 EC Q.40: solution and answer

Answer and solution

More Two-port questions