PYQs / GATE EE / 2026 / Set 1 / Q54 GATE EE 2026 Set 1 — Question 54 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MSQ +2 / -0 Medium Power Flow Equations Per-Unit System & Load Flow Power Systems
Power Systems → Per-Unit System & Load Flow → Power Flow Equations
Last updated 5 September 2026
Question 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 ∘ p.u. |V_1| = 1.1 \text{ p.u.}, |V_2| = 0.9 \text{ p.u.}, Z = 0.75 \angle 0^\circ \text{ p.u.} ∣ V 1 ∣ = 1.1 p.u. , ∣ V 2 ∣ = 0.9 p.u. , Z = 0.75∠ 0 ∘ p.u. and
θ 12 = θ 1 − θ 2 = 0 ∘ \theta_{12} = \theta_1 - \theta_2 = 0^\circ θ 12 = θ 1 − θ 2 = 0 ∘ Case-2: ∣ V 1 ∣ = 1.1 p.u. , ∣ V 2 ∣ = 0.9 p.u. , Z = 0.75 ∠ 90 ∘ p.u. |V_1| = 1.1 \text{ p.u.}, |V_2| = 0.9 \text{ p.u.}, Z = 0.75 \angle 90^\circ \text{ p.u.} ∣ V 1 ∣ = 1.1 p.u. , ∣ V 2 ∣ = 0.9 p.u. , Z = 0.75∠9 0 ∘ p.u. and
θ 12 = θ 1 − θ 2 = 90 ∘ \theta_{12} = \theta_1 - \theta_2 = 90^\circ θ 12 = θ 1 − θ 2 = 9 0 ∘ Which of the following statements is/are correct about real power loss and reactive power loss in the line?
Correct answer (A) Real 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
Solution The line current is
I = V 1 ∠ θ 12 − V 2 ∠ 0 Z ∠ θ z I = \frac{V_1 \angle \theta_{12} - V_2 \angle 0}{Z \angle \theta_z} I = Z ∠ θ z V 1 ∠ θ 12 − V 2 ∠0 .
Real power loss
P l o s s = ∣ I ∣ 2 R P_{loss} = |I|^2 R P l oss = ∣ I ∣ 2 R and reactive power loss
Q l o s s = ∣ I ∣ 2 X Q_{loss} = |I|^2 X Q l oss = ∣ I ∣ 2 X .
Case-1: Z = 0.75 ∠ 0 ∘ ⟹ R = 0.75 , X = 0 Z = 0.75 \angle 0^\circ \implies R = 0.75, X = 0 Z = 0.75∠ 0 ∘ ⟹ R = 0.75 , X = 0 .
θ 12 = 0 ∘ \theta_{12} = 0^\circ θ 12 = 0 ∘ .
∣ I 1 ∣ = ∣ 1.1 ∠ 0 ∘ − 0.9 ∠ 0 ∘ ∣ 0.75 = 0.2 0.75 = 0.2667 p.u. |I_1| = \frac{|1.1 \angle 0^\circ - 0.9 \angle 0^\circ|}{0.75} = \frac{0.2}{0.75} = 0.2667 \text{ p.u.} ∣ I 1 ∣ = 0.75 ∣1.1∠ 0 ∘ − 0.9∠ 0 ∘ ∣ = 0.75 0.2 = 0.2667 p.u. P l o s s 1 = ( 0.2667 ) 2 × 0.75 = 0.0533 p.u. P_{loss1} = (0.2667)^2 \times 0.75 = 0.0533 \text{ p.u.} P l oss 1 = ( 0.2667 ) 2 × 0.75 = 0.0533 p.u. Q l o s s 1 = ( 0.2667 ) 2 × 0 = 0 p.u. Q_{loss1} = (0.2667)^2 \times 0 = 0 \text{ p.u.} Q l oss 1 = ( 0.2667 ) 2 × 0 = 0 p.u. Case-2: Z = 0.75 ∠ 90 ∘ ⟹ R = 0 , X = 0.75 Z = 0.75 \angle 90^\circ \implies R = 0, X = 0.75 Z = 0.75∠9 0 ∘ ⟹ R = 0 , X = 0.75 .
θ 12 = 90 ∘ \theta_{12} = 90^\circ θ 12 = 9 0 ∘ .
∣ I 2 ∣ = ∣ 1.1 ∠ 90 ∘ − 0.9 ∠ 0 ∘ ∣ 0.75 = ∣ 1.1 j − 0.9 ∣ 0.75 = 1.1 2 + 0.9 2 0.75 = 1.4213 0.75 = 1.895 p.u. |I_2| = \frac{|1.1 \angle 90^\circ - 0.9 \angle 0^\circ|}{0.75} = \frac{|1.1j - 0.9|}{0.75} = \frac{\sqrt{1.1^2 + 0.9^2}}{0.75} = \frac{1.4213}{0.75} = 1.895 \text{ p.u.} ∣ I 2 ∣ = 0.75 ∣1.1∠9 0 ∘ − 0.9∠ 0 ∘ ∣ = 0.75 ∣1.1 j − 0.9∣ = 0.75 1. 1 2 + 0. 9 2 = 0.75 1.4213 = 1.895 p.u. P l o s s 2 = ( 1.895 ) 2 × 0 = 0 p.u. P_{loss2} = (1.895)^2 \times 0 = 0 \text{ p.u.} P l oss 2 = ( 1.895 ) 2 × 0 = 0 p.u. Q l o s s 2 = ( 1.895 ) 2 × 0.75 = 2.6933 p.u. Q_{loss2} = (1.895)^2 \times 0.75 = 2.6933 \text{ p.u.} Q l oss 2 = ( 1.895 ) 2 × 0.75 = 2.6933 p.u. Comparing the results:
P l o s s 1 ( 0.0533 ) > P l o s s 2 ( 0 ) ⟹ P_{loss1} (0.0533) > P_{loss2} (0) \implies P l oss 1 ( 0.0533 ) > P l oss 2 ( 0 ) ⟹ Statement (A) is correct.
Q l o s s 2 ( 2.6933 ) > Q l o s s 1 ( 0 ) ⟹ Q_{loss2} (2.6933) > Q_{loss1} (0) \implies Q l oss 2 ( 2.6933 ) > Q l oss 1 ( 0 ) ⟹ Statement (D) is correct.
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Correct answer (A) Real 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
Solution The line current is
I = V 1 ∠ θ 12 − V 2 ∠ 0 Z ∠ θ z I = \frac{V_1 \angle \theta_{12} - V_2 \angle 0}{Z \angle \theta_z} I = Z ∠ θ z V 1 ∠ θ 12 − V 2 ∠0 .
Real power loss
P l o s s = ∣ I ∣ 2 R P_{loss} = |I|^2 R P l oss = ∣ I ∣ 2 R and reactive power loss
Q l o s s = ∣ I ∣ 2 X Q_{loss} = |I|^2 X Q l oss = ∣ I ∣ 2 X .
Case-1: Z = 0.75 ∠ 0 ∘ ⟹ R = 0.75 , X = 0 Z = 0.75 \angle 0^\circ \implies R = 0.75, X = 0 Z = 0.75∠ 0 ∘ ⟹ R = 0.75 , X = 0 .
θ 12 = 0 ∘ \theta_{12} = 0^\circ θ 12 = 0 ∘ .
∣ I 1 ∣ = ∣ 1.1 ∠ 0 ∘ − 0.9 ∠ 0 ∘ ∣ 0.75 = 0.2 0.75 = 0.2667 p.u. |I_1| = \frac{|1.1 \angle 0^\circ - 0.9 \angle 0^\circ|}{0.75} = \frac{0.2}{0.75} = 0.2667 \text{ p.u.} ∣ I 1 ∣ = 0.75 ∣1.1∠ 0 ∘ − 0.9∠ 0 ∘ ∣ = 0.75 0.2 = 0.2667 p.u. P l o s s 1 = ( 0.2667 ) 2 × 0.75 = 0.0533 p.u. P_{loss1} = (0.2667)^2 \times 0.75 = 0.0533 \text{ p.u.} P l oss 1 = ( 0.2667 ) 2 × 0.75 = 0.0533 p.u. Q l o s s 1 = ( 0.2667 ) 2 × 0 = 0 p.u. Q_{loss1} = (0.2667)^2 \times 0 = 0 \text{ p.u.} Q l oss 1 = ( 0.2667 ) 2 × 0 = 0 p.u. Case-2: Z = 0.75 ∠ 90 ∘ ⟹ R = 0 , X = 0.75 Z = 0.75 \angle 90^\circ \implies R = 0, X = 0.75 Z = 0.75∠9 0 ∘ ⟹ R = 0 , X = 0.75 .
θ 12 = 90 ∘ \theta_{12} = 90^\circ θ 12 = 9 0 ∘ .
∣ I 2 ∣ = ∣ 1.1 ∠ 90 ∘ − 0.9 ∠ 0 ∘ ∣ 0.75 = ∣ 1.1 j − 0.9 ∣ 0.75 = 1.1 2 + 0.9 2 0.75 = 1.4213 0.75 = 1.895 p.u. |I_2| = \frac{|1.1 \angle 90^\circ - 0.9 \angle 0^\circ|}{0.75} = \frac{|1.1j - 0.9|}{0.75} = \frac{\sqrt{1.1^2 + 0.9^2}}{0.75} = \frac{1.4213}{0.75} = 1.895 \text{ p.u.} ∣ I 2 ∣ = 0.75 ∣1.1∠9 0 ∘ − 0.9∠ 0 ∘ ∣ = 0.75 ∣1.1 j − 0.9∣ = 0.75 1. 1 2 + 0. 9 2 = 0.75 1.4213 = 1.895 p.u. P l o s s 2 = ( 1.895 ) 2 × 0 = 0 p.u. P_{loss2} = (1.895)^2 \times 0 = 0 \text{ p.u.} P l oss 2 = ( 1.895 ) 2 × 0 = 0 p.u. Q l o s s 2 = ( 1.895 ) 2 × 0.75 = 2.6933 p.u. Q_{loss2} = (1.895)^2 \times 0.75 = 2.6933 \text{ p.u.} Q l oss 2 = ( 1.895 ) 2 × 0.75 = 2.6933 p.u. Comparing the results:
P l o s s 1 ( 0.0533 ) > P l o s s 2 ( 0 ) ⟹ P_{loss1} (0.0533) > P_{loss2} (0) \implies P l oss 1 ( 0.0533 ) > P l oss 2 ( 0 ) ⟹ Statement (A) is correct.
Q l o s s 2 ( 2.6933 ) > Q l o s s 1 ( 0 ) ⟹ Q_{loss2} (2.6933) > Q_{loss1} (0) \implies Q l oss 2 ( 2.6933 ) > Q l oss 1 ( 0 ) ⟹ Statement (D) is correct.
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Case-1: and