GATE 2026 IN Q.39: solution and answer

GATE 2026 · INQ.39 · 2 marks · MCQ

A two-port network has z-parameters $z_{11} = z_{22} = 10 Ω$ , and $z_{12} = z_{21} = 5 Ω$ .

The value of $R_{L}$ such that maximum power is transferred to $R_{L}$ is ________ $Ω$ .

GATE 2026 IN question 39: A two-port network is drawn as a box labelled with the matrix entries z11 z12 (top row) and z21 z22 (bottom row). — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
5.67
(B)
8.33
(C)
10.33
(D)
25.67

Answer and solution

ANSWER

$R_{L} = Z_{t h} = z_{22} - \frac{z _{12} z _{21}}{z _{11} + R _{s}} = 10 - \frac{25}{15} = \frac{25}{3} \approx 8.33 Ω$ , option (B).

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Port equations from the z-parameters, with both port currents taken as flowing into the network: $V_{1} = z_{11} I_{1} + z_{12} I_{2} = 10 I_{1} + 5 I_{2}$ and $V_{2} = z_{21} I_{1} + z_{22} I_{2} = 5 I_{1} + 10 I_{2}$ .
Source constraint at port 1 (KVL around the 5 V source and the $5 Ω$ resistor): $V_{1} = 5 - 5 I_{1}$ .
To find the Thevenin impedance seen by $R_{L}$ , kill the 5 V source (replace it with a short) and drive port 2 with a test source $V_{2}$ that pushes current $I_{2}$ into the network. Port 1 is now loaded only by the $5 Ω$ resistor, so $V_{1} = - 5 I_{1}$ .
Substitute into the first port equation: $- 5 I_{1} = 10 I_{1} + 5 I_{2}$ , so $- 15 I_{1} = 5 I_{2}$ and $I_{1} = - \frac{I _{2}}{3}$ .
Substitute $I_{1}$ into the second port equation: $V_{2} = 5 (- \frac{I _{2}}{3}) + 10 I_{2} = \frac{25}{3} I_{2}$ . Hence $Z_{t h} = \frac{V _{2}}{I _{2}} = z_{22} - \frac{z _{12} z _{21}}{z _{11} + R _{s}} = 10 - \frac{25}{15} = \frac{25}{3} Ω.$
Maximum power transfer theorem: with a purely resistive Thevenin impedance, a resistive load draws maximum power when $R_{L} = Z_{t h} = \frac{25}{3} \approx 8.33 Ω$ , which is option (B). The open-circuit voltage, $V_{t h} = z_{21} I_{1}_{I_{2} = 0} = 5 \cdot \frac{5}{10 + 5} = \frac{5}{3} V$ , sets how much power reaches the load but does not change the value of $R_{L}$ that maximises it.

KEY IDEA

Set $R_{L}$ equal to the Thevenin impedance at port 2. That impedance is the network's output impedance with the source's own $5 Ω$ resistance ( $R_{s}$ ) loading port 1: $Z_{o u t} = z_{22} - \frac{z _{12} z _{21}}{z _{11} + R _{s}}$ .

WATCH OUT FOR

Leaving out the source resistance: $z_{22} - \frac{z _{12} z _{21}}{z _{11}} = 10 - 2.5 = 7.5 Ω$ . Taking $R_{L} = z_{22} = 10 Ω$ , which is the output impedance only when port 1 is open. Adding the correction term instead of subtracting it, which gives $11.67 Ω$ . Another mistake is killing the 5 V source by open-circuiting it; a voltage source is killed by shorting it, so the $5 Ω$ resistor stays connected across port 1.

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2026 IN Q.39: solution and answer

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