GATE 2026 EE Q.60: solution and answer

GATE 2026 · EEQ.60 · 2 marks · NAT

Consider the two-port network shown. For maximum power transfer to the resistive load ( $R_{L}$ ), the value of $R_{L}$ should be ________ $Ω$ (Round off to two decimal places)

GATE 2026 EE question 60: An ideal 10 V DC voltage source, + terminal on top, drives port 1; — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

ANSWER

For maximum power transfer, $R_{L} = R_{t h} = 10 ∥ 4 = \frac{20}{7} \approx 2.86 Ω$ .

Official answer: 2.81 to 2.91Source: Official answer key, GATE 2026 EE (IIT Guwahati)

Maximum power transfer theorem: a resistive load draws maximum power when $R_{L} = R_{t h}$ , the Thevenin resistance seen from port 2 (the terminals of $R_{L}$ ).
To find $R_{t h}$ , deactivate the independent source: the ideal 10 V source is replaced by a short circuit between the top-left node and the bottom rail.
That short sits directly across the 5 Ω shunt resistor, so the 5 Ω carries no current and drops out. The 10 Ω series resistor, through the short, now connects the port-2 top node to the bottom rail.
Seen from port 2, the 10 Ω path is therefore in parallel with the 4 Ω shunt: $R_{t h} = 10 ∥ 4 = \frac{10 \times 4}{10 + 4} = \frac{40}{14} = 2.857 Ω$
Check (not needed for the answer): with $R_{L}$ removed, the 5 Ω sits directly across the ideal source and does not affect the rest, so $V_{t h} = 10 \times \frac{4}{10 + 4} = 2.857 V$ and $P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{( 2.857 ) ^{2}}{4 \times 2.857} = 0.714 W$ .
So $R_{L} = R_{t h} = \frac{20}{7} \approx 2.86 Ω$ .

KEY IDEA

Match $R_{L}$ to the Thevenin resistance at port 2. When the ideal voltage source is replaced by a short, it shorts out the 5 Ω shunt at port 1. That leaves the 10 Ω series arm in parallel with the 4 Ω shunt at port 2.

WATCH OUT FOR

Open-circuiting the voltage source instead of shorting it keeps the 5 Ω in the path and gives $(10 + 5) ∥ 4 = 60/19 \approx 3.16 Ω$ , which is wrong. Another mistake is working out two-port parameters (Z or Y) of the pi network alone. That ignores the fact that the ideal source fixes $v_{1}$ and so shorts port 1 when it is deactivated.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2026 EE Q.60: solution and answer

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