60 solved GATE EE questions on Power Systems, drawn from 10 exam years and grouped by year. Every question shows the official answer and a step-by-step solution.
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Declare every quantity's reference: phase or line, total or per-phase, physical or per-unit. These selected notes develop balanced steady-state reasoning and base conversion through original analytical exercises, not installation procedures.
Our study notes and original examples support the PYQs below; they are not official exam questions or a replacement for the current syllabus.
With identical balanced phases, total complex power is S₃φ = 3Vphase Iphase*. For line RMS magnitudes, P = √3 VLL IL cosφ and Q = √3 VLL IL sinφ. The angle φ describes the corresponding phase voltage-current displacement, not an arbitrary line-voltage phasor angle.
Check yourself: Are your voltage and current a matched phase pair?
Choose total three-phase apparent-power base Sbase and line-to-line RMS voltage base Vbase. Then Ibase = Sbase/(√3 Vbase) and Zbase = Vbase²/Sbase. The corresponding phase-voltage base is Vbase/√3. These definitions remove the extra three-phase factors from normalized equations.
Check yourself: Did you accidentally combine total power with a phase-voltage base?
Convert using Zpu,new = Zpu,old × (Snew/Sold) × (Vold/Vnew)². This follows by recovering the same physical impedance from either base. Across an ideal transformer, select voltage bases following its rated ratio when simplifying the per-unit network.
Check yourself: Does converting back to ohms give the same original impedance?
The admittance matrix relates bus-current injections to bus voltages. Complex-power injection uses voltage times conjugate current. Consuming loads have negative injection under a generation-positive convention; load-absorption calculations use the opposite reference.
Check yourself: Is positive power defined as entering the network or entering the load?
A balanced three-phase fault can be analyzed with the positive-sequence network under the usual symmetrical model. Unbalanced faults generally require sequence-network interconnections determined by fault conditions. Zero-sequence paths depend on the specified network and transformer connections, not simply on positive-sequence impedance.
Check yourself: Does the stated fault preserve balance or require other sequence components?
Original teaching example · not a PYQ
Original practice example: a balanced wye load absorbs total S = 12 + j9 MVA at 20 kV line-to-line RMS in sinusoidal steady state. Use Sbase = 25 MVA three-phase and Vbase = 20 kV line-to-line. Ignore feeder impedance and take phase-a load voltage as angle zero. Find line-current magnitude in per-unit and amperes.
Line-current magnitude = 0.6 pu = 250√3 A, approximately 433.013 A RMS.
Separately, a reactance is 0.2 pu on 50 MVA three-phase and 10 kV line-to-line bases. At the same voltage location, express the unchanged reactance on 100 MVA and 20 kV bases.
Reactance = 0.1 pu on the new bases.
The conversion gives 0.2 × (100/50) × (10/20)² = 0.1. Independently, the old impedance base is 2 Ω and the new base is 4 Ω. Both representations therefore describe the same 0.4 Ω reactance.
Supplemental reading, not an official GATE reading list or an endorsement of these notes.
Free supplemental chapter on balanced polyphase networks, transformers, and per-unit representation. Re-derive the current and impedance bases before reading further.
This page shows 60 recent questions from the released archive, newest first. For older questions and complete papers, browse all GATE EE papers. Questions can carry more than one subject tag; counts are not marks weightage.
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