PYQs / GATE EE / 2014 / Set 1 / Q51 GATE EE 2014 Set 1 — Question 51 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 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 A two bus power system shown in the figure supplies load of
1.0 + j 0.5 1.0+j0.5 1.0 + j 0.5 p.u.
The values of
V 1 V_1 V 1 in p.u. and
δ 2 \delta_2 δ 2 respectively are
Correct answer (B) 1.05 and -5.44^
Solution The current flowing from Bus 1 to Bus 2 is
I 12 = V 1 ∠ 0 ∘ − 1 ∠ δ 2 j 0.1 I_{12} = \frac{V_1 \angle 0^\circ - 1 \angle \delta_2}{j0.1} I 12 = j 0.1 V 1 ∠ 0 ∘ − 1∠ δ 2 .
The load current at Bus 2 is
I L = ( S L V 2 ) ∗ = ( 1 + j 0.5 1 ∠ δ 2 ) ∗ = ( 1 − j 0.5 ) ∠ δ 2 I_L = \left(\frac{S_L}{V_2}\right)^* = \left(\frac{1+j0.5}{1 \angle \delta_2}\right)^* = (1-j0.5) \angle \delta_2 I L = ( V 2 S L ) ∗ = ( 1∠ δ 2 1 + j 0.5 ) ∗ = ( 1 − j 0.5 ) ∠ δ 2 .
Equating
I 12 = I L I_{12} = I_L I 12 = I L :
V 1 − ( cos δ 2 + j sin δ 2 ) j 0.1 = ( 1 − j 0.5 ) ( cos δ 2 + j sin δ 2 ) \frac{V_1 - (\cos \delta_2 + j \sin \delta_2)}{j0.1} = (1-j0.5)(\cos \delta_2 + j \sin \delta_2) j 0.1 V 1 − ( c o s δ 2 + j s i n δ 2 ) = ( 1 − j 0.5 ) ( cos δ 2 + j sin δ 2 ) V 1 − cos δ 2 − j sin δ 2 = j 0.1 ( cos δ 2 + j sin δ 2 − j 0.5 cos δ 2 + 0.5 sin δ 2 ) V_1 - \cos \delta_2 - j \sin \delta_2 = j0.1(\cos \delta_2 + j \sin \delta_2 - j0.5 \cos \delta_2 + 0.5 \sin \delta_2) V 1 − cos δ 2 − j sin δ 2 = j 0.1 ( cos δ 2 + j sin δ 2 − j 0.5 cos δ 2 + 0.5 sin δ 2 ) V 1 − cos δ 2 − j sin δ 2 = j 0.1 cos δ 2 − 0.1 sin δ 2 + 0.05 cos δ 2 + j 0.05 sin δ 2 V_1 - \cos \delta_2 - j \sin \delta_2 = j0.1 \cos \delta_2 - 0.1 \sin \delta_2 + 0.05 \cos \delta_2 + j0.05 \sin \delta_2 V 1 − cos δ 2 − j sin δ 2 = j 0.1 cos δ 2 − 0.1 sin δ 2 + 0.05 cos δ 2 + j 0.05 sin δ 2 Equating imaginary parts:
− sin δ 2 = 0.1 cos δ 2 + 0.05 sin δ 2 ⇒ − 1.05 sin δ 2 = 0.1 cos δ 2 -\sin \delta_2 = 0.1 \cos \delta_2 + 0.05 \sin \delta_2 \Rightarrow -1.05 \sin \delta_2 = 0.1 \cos \delta_2 − sin δ 2 = 0.1 cos δ 2 + 0.05 sin δ 2 ⇒ − 1.05 sin δ 2 = 0.1 cos δ 2 tan δ 2 = − 0.1 1.05 ⇒ δ 2 = arctan ( − 0.0952 ) ≈ − 5.44 ∘ \tan \delta_2 = -\frac{0.1}{1.05} \Rightarrow \delta_2 = \arctan(-0.0952) \approx -5.44^\circ tan δ 2 = − 1.05 0.1 ⇒ δ 2 = arctan ( − 0.0952 ) ≈ − 5.4 4 ∘ Equating real parts:
V 1 − cos δ 2 = − 0.1 sin δ 2 + 0.05 cos δ 2 ⇒ V 1 = 1.05 cos δ 2 − 0.1 sin δ 2 V_1 - \cos \delta_2 = -0.1 \sin \delta_2 + 0.05 \cos \delta_2 \Rightarrow V_1 = 1.05 \cos \delta_2 - 0.1 \sin \delta_2 V 1 − cos δ 2 = − 0.1 sin δ 2 + 0.05 cos δ 2 ⇒ V 1 = 1.05 cos δ 2 − 0.1 sin δ 2 V 1 = 1.05 cos ( − 5.44 ∘ ) − 0.1 sin ( − 5.44 ∘ ) ≈ 1.05 ( 0.9955 ) + 0.1 ( 0.0948 ) ≈ 1.054 V_1 = 1.05 \cos(-5.44^\circ) - 0.1 \sin(-5.44^\circ) \approx 1.05(0.9955) + 0.1(0.0948) \approx 1.054 V 1 = 1.05 cos ( − 5.4 4 ∘ ) − 0.1 sin ( − 5.4 4 ∘ ) ≈ 1.05 ( 0.9955 ) + 0.1 ( 0.0948 ) ≈ 1.054 p.u.
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Correct answer (B) 1.05 and -5.44^
Solution The current flowing from Bus 1 to Bus 2 is
I 12 = V 1 ∠ 0 ∘ − 1 ∠ δ 2 j 0.1 I_{12} = \frac{V_1 \angle 0^\circ - 1 \angle \delta_2}{j0.1} I 12 = j 0.1 V 1 ∠ 0 ∘ − 1∠ δ 2 .
The load current at Bus 2 is
I L = ( S L V 2 ) ∗ = ( 1 + j 0.5 1 ∠ δ 2 ) ∗ = ( 1 − j 0.5 ) ∠ δ 2 I_L = \left(\frac{S_L}{V_2}\right)^* = \left(\frac{1+j0.5}{1 \angle \delta_2}\right)^* = (1-j0.5) \angle \delta_2 I L = ( V 2 S L ) ∗ = ( 1∠ δ 2 1 + j 0.5 ) ∗ = ( 1 − j 0.5 ) ∠ δ 2 .
Equating
I 12 = I L I_{12} = I_L I 12 = I L :
V 1 − ( cos δ 2 + j sin δ 2 ) j 0.1 = ( 1 − j 0.5 ) ( cos δ 2 + j sin δ 2 ) \frac{V_1 - (\cos \delta_2 + j \sin \delta_2)}{j0.1} = (1-j0.5)(\cos \delta_2 + j \sin \delta_2) j 0.1 V 1 − ( c o s δ 2 + j s i n δ 2 ) = ( 1 − j 0.5 ) ( cos δ 2 + j sin δ 2 ) V 1 − cos δ 2 − j sin δ 2 = j 0.1 ( cos δ 2 + j sin δ 2 − j 0.5 cos δ 2 + 0.5 sin δ 2 ) V_1 - \cos \delta_2 - j \sin \delta_2 = j0.1(\cos \delta_2 + j \sin \delta_2 - j0.5 \cos \delta_2 + 0.5 \sin \delta_2) V 1 − cos δ 2 − j sin δ 2 = j 0.1 ( cos δ 2 + j sin δ 2 − j 0.5 cos δ 2 + 0.5 sin δ 2 ) V 1 − cos δ 2 − j sin δ 2 = j 0.1 cos δ 2 − 0.1 sin δ 2 + 0.05 cos δ 2 + j 0.05 sin δ 2 V_1 - \cos \delta_2 - j \sin \delta_2 = j0.1 \cos \delta_2 - 0.1 \sin \delta_2 + 0.05 \cos \delta_2 + j0.05 \sin \delta_2 V 1 − cos δ 2 − j sin δ 2 = j 0.1 cos δ 2 − 0.1 sin δ 2 + 0.05 cos δ 2 + j 0.05 sin δ 2 Equating imaginary parts:
− sin δ 2 = 0.1 cos δ 2 + 0.05 sin δ 2 ⇒ − 1.05 sin δ 2 = 0.1 cos δ 2 -\sin \delta_2 = 0.1 \cos \delta_2 + 0.05 \sin \delta_2 \Rightarrow -1.05 \sin \delta_2 = 0.1 \cos \delta_2 − sin δ 2 = 0.1 cos δ 2 + 0.05 sin δ 2 ⇒ − 1.05 sin δ 2 = 0.1 cos δ 2 tan δ 2 = − 0.1 1.05 ⇒ δ 2 = arctan ( − 0.0952 ) ≈ − 5.44 ∘ \tan \delta_2 = -\frac{0.1}{1.05} \Rightarrow \delta_2 = \arctan(-0.0952) \approx -5.44^\circ tan δ 2 = − 1.05 0.1 ⇒ δ 2 = arctan ( − 0.0952 ) ≈ − 5.4 4 ∘ Equating real parts:
V 1 − cos δ 2 = − 0.1 sin δ 2 + 0.05 cos δ 2 ⇒ V 1 = 1.05 cos δ 2 − 0.1 sin δ 2 V_1 - \cos \delta_2 = -0.1 \sin \delta_2 + 0.05 \cos \delta_2 \Rightarrow V_1 = 1.05 \cos \delta_2 - 0.1 \sin \delta_2 V 1 − cos δ 2 = − 0.1 sin δ 2 + 0.05 cos δ 2 ⇒ V 1 = 1.05 cos δ 2 − 0.1 sin δ 2 V 1 = 1.05 cos ( − 5.44 ∘ ) − 0.1 sin ( − 5.44 ∘ ) ≈ 1.05 ( 0.9955 ) + 0.1 ( 0.0948 ) ≈ 1.054 V_1 = 1.05 \cos(-5.44^\circ) - 0.1 \sin(-5.44^\circ) \approx 1.05(0.9955) + 0.1(0.0948) \approx 1.054 V 1 = 1.05 cos ( − 5.4 4 ∘ ) − 0.1 sin ( − 5.4 4 ∘ ) ≈ 1.05 ( 0.9955 ) + 0.1 ( 0.0948 ) ≈ 1.054 p.u.
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