📘 Power Systems – Full Length Mock Test (Page 19)
Time: 60 Minutes | Total Questions: 25 Mixed: Concept + Numerical + MCQ
🔹 Section A – Concept MCQ (1–10)
1. Most severe fault is: A) SLG B) LL C) LLG D) 3-phase 2. Per unit impedance remains constant across transformer when base is properly chosen. (True/False) 3. Surge impedance = A) √(R/L) B) √(L/C) C) √(C/L) D) L/C 4. Equal area criterion applies to: A) Voltage stability B) Rotor angle stability C) Frequency control D) Economic dispatch 5. NR method convergence is: A) Linear B) Quadratic C) Exponential D) Slow 6. Penalty factor used when: A) No losses B) Transmission losses present C) Voltage high D) Load low 7. Maximum steady-state power occurs at δ = ? 8. Slack bus compensates for: A) Voltage B) Reactive power C) Losses D) Frequency 9. Corona increases with: A) Voltage B) Diameter C) Spacing D) Height 10. HVDC eliminates: A) Reactive power flow B) Resistance C) Losses D) Voltage drop🔹 Section B – Numerical (11–20)
11. Generator E = 1.1 pu, V = 1 pu, X = 0.5 pu Find Pmax. 12. A 220 kV bus has fault level 6000 MVA Find short circuit current. 13. Two generators: X1 = 0.2 pu, X2 = 0.3 pu Find equivalent reactance. 14. Base MVA = 100, Base kV = 11 Find base current. 15. If impedance reduces by 25%, fault current changes by? 16. Economic dispatch: dC1/dP1 = 0.04P1 + 4 dC2/dP2 = 0.02P2 + 5 Total load = 200 MW Find P1 and P2. 17. Line reactance = j0.4 pu Load = 1 + j0.5 pu Find approximate current magnitude. 18. If inertia constant doubles, stability margin: A) Decreases B) Increases C) Same D) Zero 19. Fault current = 2 pu System base = 100 MVA, 132 kV Find actual current. 20. Load factor = 0.7 Peak load = 100 MW Find average load.🔹 Section C – Advanced (21–25)
21. Derive condition for maximum power transfer in SMIB system. 22. Why does distance relay underreach due to infeed? 23. Explain importance of penalty factor in dispatch. 24. How does increase in reactance affect stability? 25. Compare NR and Gauss-Seidel load flow methods.🔹 Answers
1. D 2. True 3. B 4. B 5. B 6. B 7. 90° 8. C 9. A 10. A 11. 2.2 pu 12. I = 6000 / (1.732 × 220) = 15.7 kA 13. Xeq = 0.12 pu 14. Ib = S/(√3V) = 100/(1.732×11) ≈ 5.25 kA 15. Increases by 33% 16. Solve → P1 ≈ 80 MW, P2 ≈ 120 MW 17. ≈ 1.12 pu 18. B 19. Iactual = (2 × 100)/(1.732×132) ≈ 0.87 kA 20. 70 MWPage 19 – Full Length Mock Test Completed
.png)
No comments:
Post a Comment