GATE 2026 EE Q.54: solution and answer

GATE 2026 · EEQ.54 · 2 marks · MSQ

For the balanced 3-phase transmission line shown, consider the following cases: Case-1: $∣ V_{1} ∣ = 1.1$ p.u., $∣ V_{2} ∣ = 0.9$ p.u., $Z = 0.75∠ 0^{\circ}$ p.u. and $θ_{12} = θ_{1} - θ_{2} = 0^{\circ}$ . Case-2: $∣ V_{1} ∣ = 1.1$ p.u., $∣ V_{2} ∣ = 0.9$ p.u., $Z = 0.75∠9 0^{\circ}$ p.u. and $θ_{12} = θ_{1} - θ_{2} = 9 0^{\circ}$ . Which of the following statements is/are correct about real power loss and reactive power loss in the line?

GATE 2026 EE question 54: Single-line diagram: generator G1 (circle) connected to a bus on the left with voltage V1∠θ1, a horizontal transmission… — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
Real power loss in Case-1 is more than that in Case-2
(B)
Real power loss in Case-2 is more than that in Case-1
(C)
Reactive power loss in Case-1 is more than that in Case-2
(D)
Reactive power loss in Case-2 is more than that in Case-1

Answer and solution

ANSWER

Options (A) and (D) are correct.

Official answer: A, DSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Losses in a series impedance $Z = R + j X$ carrying current $I$ are $S_{l oss} = ∣ I ∣^{2} Z$ . So $P_{l oss} = ∣ I ∣^{2} R$ and $Q_{l oss} = ∣ I ∣^{2} X$ , with $I = \frac{V _{1} - V _{2}}{Z}$ (all values in p.u.).
Case-1: $Z = 0.75∠ 0^{\circ} = 0.75 + j 0$ , so $R = 0.75$ and $X = 0$ . Also $V_{1} = 1.1∠ 0^{\circ}$ and $V_{2} = 0.9∠ 0^{\circ}$ (since $θ_{12} = 0^{\circ}$ ), so $V_{1} - V_{2} = 0.2$ .
Case-1 current: $∣ I ∣ = \frac{0.2}{0.75} = 0.267$ p.u., so $∣ I ∣^{2} = 0.0711$ .
Case-1 losses: $P_{l oss} = 0.0711 \times 0.75 \approx 0.053$ p.u. and $Q_{l oss} = 0.0711 \times 0 = 0$ .
Case-2: $Z = 0.75∠9 0^{\circ} = j 0.75$ , so $R = 0$ and $X = 0.75$ . Taking $θ_{2} = 0^{\circ}$ , $θ_{1} = 9 0^{\circ}$ : $V_{1} = 1.1∠9 0^{\circ} = j 1.1$ and $V_{2} = 0.9∠ 0^{\circ} = 0.9$ .
Case-2 voltage drop: $V_{1} - V_{2} = - 0.9 + j 1.1$ , so $∣ V_{1} - V_{2} ∣ = 0.81 + 1.21 = 2.02 = 1.421$ p.u.
Case-2 current: $∣ I ∣ = \frac{1.421}{0.75} = 1.895$ p.u., so $∣ I ∣^{2} = 3.591$ .
Case-2 losses: $P_{l oss} = 3.591 \times 0 = 0$ and $Q_{l oss} = 3.591 \times 0.75 \approx 2.69$ p.u.
Compare: real loss is $0.053$ in Case-1 and $0$ in Case-2, so (A) is true and (B) is false. Reactive loss is $0$ in Case-1 and $2.69$ in Case-2, so (C) is false and (D) is true. Correct options: A and D.

KEY IDEA

Real loss is $∣ I ∣^{2} R$ and reactive loss is $∣ I ∣^{2} X$ , so the angle of $Z$ decides which kind of loss the line has. Find $I$ from the voltage difference first, then split $Z$ into $R$ and $X$ .

WATCH OUT FOR

Using $∣ V_{1} ∣ - ∣ V_{2} ∣$ instead of the phasor difference $V_{1} - V_{2}$ in Case-2. The $9 0^{\circ}$ angle between the voltages makes the difference $1.42$ p.u., not $0.2$ p.u. Another slip is treating a purely reactive line as lossless: it has no real loss but it absorbs reactive power.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 6 Transmission & distribution questions, with notes

GATE 2026 EE Q.54: solution and answer

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