📘 Power Systems – Question Bank (Page 14)
🔹 Problem 1: Multi-Source Fault Calculation
Two generators feed a 220 kV bus through reactances:- G1: X = 0.2 pu
- G2: X = 0.25 pu
- Line reactance to fault point = 0.3 pu
Parallel generator reactance: 1/Xeq = 1/0.2 + 1/0.25 = 5 + 4 = 9 Xeq = 1/9 = 0.111 pu Total reactance: Ztotal = 0.111 + 0.3 = 0.411 pu Fault current = 1 / 0.411 = 2.43 pu
🔹 Problem 2: Critical Clearing Angle
A generator has:- Pmax = 2 pu
- Mechanical input Pm = 1 pu
Pm = Pmax sinδ₀ 1 = 2 sinδ₀ sinδ₀ = 0.5 δ₀ = 30°
🔹 Problem 3: Economic Dispatch with Transmission Loss
Loss formula: PL = 0.0005P1² Total demand = 300 MW Generators: C1 = 0.02P1² + 3P1 C2 = 0.025P2² + 2P2Condition: dC1/dP1 × Penalty1 = dC2/dP2 × Penalty2 Penalty factor ≈ 1 / (1 - dPL/dP1) dPL/dP1 = 0.001P1 Complex iterative solution required (Advanced level problem – lambda iteration used)
🔹 Problem 4: Load Flow – Jacobian Matrix Insight
Explain why NR converges faster.Because Newton-Raphson uses first-order Taylor expansion and updates voltage magnitude and angle simultaneously. Convergence is quadratic.
🔹 Problem 5: HVDC Power Control
An HVDC link operates at 400 kV and 1000 A. If voltage increases to 450 kV, find percentage power increase.Initial Power = 400 × 1000 = 400 MW New Power = 450 × 1000 = 450 MW Increase = 50 MW Percentage increase = (50/400) × 100 = 12.5%
🔹 Problem 6: Distance Relay Reach
Line impedance = 50 Ω Zone 1 setting = 80% Zone 2 setting = 120%Zone 1 reach = 0.8 × 50 = 40 Ω Zone 2 reach = 1.2 × 50 = 60 Ω
🔹 Final Challenge
If system impedance reduces by 20%, how does fault current change?I = V / Z If Z reduces by 20% → Znew = 0.8Z Inew = V / (0.8Z) = 1.25 I Fault current increases by 25%
Page 14 – Very Hard GATE-Level Numerical Completed
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