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Electrical Engineering 2025 Paper II 50 marks Solve

Paper II — Q6

(a) Determine the sending-end voltage, current, power factor of a single-phase, 50 Hz, 76·2 kV transmission line delivering a…

(a)

Determine the sending-end voltage, current, power factor of a single-phase, 50 Hz, 76·2 kV transmission line delivering a load of 12 MW at 0·8 p.f. lagging. The line constants are R = 25 Ω, inductance 200 mH and capacitance between lines is 2·5 μF. Also determine the regulation and efficiency of the transmission. Use nominal-π method. Draw the phasor diagram. 20 marks

(b)

24 voice signals are sampled uniformly and then time-division multiplexed. Flat-top sampling is used with one microsecond duration. Multiplexing operation provides for synchronization by adding an extra pulse of sufficient amplitude and also one microsecond duration. The highest frequency component of each voice signal is 3·4 kHz. (i) Assuming a sampling rate of 8 kHz, find the spacing between successive pulses of the multiplexed signal. (ii) Repeat your calculations by assuming the use of Nyquist rate sampling. 20 marks

(c)

A 3-phase, 33 kV, star-connected alternator is to be protected using circulating current protection. The pilot wires are connected to the secondary windings of 100/5 ratio current transformer. The protective relay is adjusted to operate with an out of balance current of 1 A in the pilot wires. Determine (i) the earthing resistance which will protect 90% of the winding and (ii) the percent of the winding which would be protected, if the earthing resistance is 15 Ω. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

एक एकल-कला, 50 Hz, 76·2 kV संचरण लाइन के प्रेषण-छोर पर वोल्टता, धारा तथा शक्ति गुणक का निर्धारण कीजिये, जबकि संचरण लाइन 12 MW का भार, 0·8 पश्चगामी शक्ति गुणक पर प्रदान करती है। संचरण लाइन के नियतांक R = 25 Ω, प्रेरकत्व 200 mH तथा लाइनों के मध्य धारिता 2·5 μF है। संचरण की विनियमता (रियुलेशन) एवं दक्षता का भी निर्धारण कीजिये। सांकेतिक-π विधि का प्रयोग कीजिये। कला आरेख (फेजर डायग्राम) को आरेखित कीजिये। 20 अंक

(b)

24 ध्वनि संकेतों को एकसमान रूप से पहले प्रतिचयनित (सैम्पल्ड) किया जाता है, उसके उपरान्त उन्हें काल-विभाजन (टाइम-डिविजन) से बहुल (मल्टीप्लेक्स्ड) किया जाता है। एक माइक्रोसेकंड अवधि के समतल-शीर्ष (फ्लैट-टॉप) प्रतिचयन का उपयोग किया जाता है। बहुल संचालन, पर्याप्त आयाम एवं एक माइक्रोसेकंड अवधि की एक अतिरिक्त पल्स संयोजित कर तुल्यकालन (सिंक्रोनाइजेशन) प्रदान करता है। प्रत्येक ध्वनि संकेत का उच्चतम आवृत्ति घटक 3·4 kHz है। (i) प्रतिचयन दर को 8 kHz मानते हुए बहुल संकेत के उत्तरोत्तर पल्स के मध्य अन्तराल प्राप्त कीजिये। (ii) नाइक्विस्ट प्रतिचयन दर का उपयोग मानते हुए अपनी गणना को दोहराइये। 20 अंक

(c)

एक 3-कला, 33 kV, तारा (स्टार)-संयोजित प्रत्यावर्ती धारा जनित्र को परिसंचारी धारा संरक्षण का उपयोग कर संरक्षित किया जाता है। सूचक तारों (पाइलट वायर) को 100/5 अनुपात के धारा परिणामित्र (करेंट ट्रांसफॉर्मर) की द्वितीयक कुंडली से संयोजित किया गया है। संरक्षी रिले का संचालन, सूचक तारों में 1 A की असंतुलित धारा के लिये समायोजित किया गया है। निर्धारण कीजिये (i) भूसंपर्कन प्रतिरोध (अर्थिंग रेजिस्टेंस) का मान, जो कि 90% कुंडली को संरक्षित करेगा और (ii) संरक्षित कुंडली का प्रतिशत, यदि भूसंपर्कन प्रतिरोध 15 Ω है। 10 अंक

Q6 of the 2025 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2025 Electrical Engineering paper
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Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a)

Take receiving-end voltage as reference: V_R = 76.2 kV ∠0° = 76200 V. Load: P = 12 MW, p.f. = 0.8 lagging, so φ_R = cos⁻¹ 0.8 = 36.87°. I_R = P/(V_R cos φ_R) = 12×10⁶/(76200×0.8) = 196.85 A. I_R = 196.85∠−36.87° = 157.48 − j118.11 A.

Line constants: R = 25 Ω, X_L = 2π×50×0.2 = 62.8319 Ω. Z = 25 + j62.8319 Ω. C = 2.5 μF, Y = j2π×50×2.5×10⁻⁶ = j0.000785398 S, so Y/2 = j0.000392699 S.

By nominal-π method, receiving-end capacitor current: I_C_R = V_R(Y/2) = j29.9237 A. Series line current: I_line = I_R + I_C_R = 157.480 − j88.187 A. I_line Z = 9477.93 + j7690.12 V. V_S = V_R + I_line Z = 85677.93 + j7690.12 V. |V_S| = 86022.36 V ≈ 86.022 kV.

Sending-end capacitor current: I_C_S = V_S(Y/2) = −3.0199 + j33.6456 A. I_S = I_line + I_C_S = 154.460 − j54.541 A. |I_S| = 163.807 A.

Angle of V_S = 5.13°, angle of I_S = −19.45°. Sending-end power-factor angle = 5.13° − (−19.45°) = 24.58°. Sending-end p.f. = cos 24.58° = 0.909 lagging.

For regulation, the no-load receiving voltage with the same sending voltage is V_R(no load) = |V_S|/|A|, where A = 1 + ZY/2. ZY/2 = −0.024674 + j0.0098175, so |A| = 0.975375. V_R(no load) = 86022.36/0.975375 = 88194.1 V. Regulation = (88194.1 − 76200)/76200 × 100 = 15.74%.

Copper loss = |I_line|²R = (180.491)²×25 = 814423 W = 0.8144 MW. Efficiency = 12/(12 + 0.8144) × 100 = 93.65%.

Phasor diagram (described): Take V_R along reference. I_R lags V_R by 36.87°. I_C_R leads V_R by 90°. I_line = I_R + I_C_R. V_S = V_R + I_lineZ, leading V_R by 5.13°. I_C_S leads V_S by 90°, and I_S = I_line + I_C_S lags V_S by 24.58°.

Diagram: Nominal-pi phasor diagram: V_R reference, I_R lags 36.87 degrees, V_S = 86.02 kV leads by 5.13 degrees, I_S = 163.8 A lags V_S by 24.58 degr

(b)

Number of pulses per frame including synchronizing pulse = 24 + 1 = 25. Pulse duration τ = 1 μs.

(i) Sampling rate = 8 kHz. T_s = 1/8000 = 125 μs. Slot width = T_s/25 = 125/25 = 5 μs. Clear spacing between successive pulses = slot width − τ = 5 − 1 = 4 μs. Leading-edge-to-leading-edge spacing = 5 μs.

(ii) Nyquist rate = 2×3.4 kHz = 6.8 kHz. T_s = 1/6800 = 147.0588 μs. Slot width = 147.0588/25 = 5.88235 μs. Clear spacing = 5.88235 − 1 = 4.88235 μs. Leading-edge-to-leading-edge spacing = 5.88235 μs.

(c)

CT ratio = 100/5 = 20. Relay operating current = 1 A in pilot wires. Primary out-of-balance current I_th = 1×20 = 20 A.

Phase voltage V_ph = 33000/√3 = 19052.56 V. For a fault at fraction α from the neutral end, I_f = αV_ph/R_e, neglecting winding impedance and fault resistance.

(i) To protect 90% of winding, the 10% nearest neutral may remain unprotected, so α = 0.10. R_e = αV_ph/I_th = 0.10×19052.56/20. R_e = 95.26 Ω (or less, to protect at least 90%).

(ii) If R_e = 15 Ω, α = I_thR_e/V_ph = 20×15/19052.56 = 0.01575. Unprotected winding near neutral = 1.57%. Protected winding = 100 − 1.57 = 98.43%.

How this answer will be evaluated

What earns full marks: Complete equivalent circuits, accurate phasor diagrams, all calculations with units, correct final values with proper interpretation

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Determine sending-end voltage, current, p.f., regulation, and efficiency using nominal-π method. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Nominal-π: Z=25+j62.83 Ω; Y/2=j3.927×10⁻⁴ S
    • VR=76200∠0°; IR=196.85∠−36.87° A; I_line=IR+IC_R=157.48−j88.19 A
    • VS=VR+I_lineZ ≈ 86.02 kV; IS=163.81 A; sending pf 0.909 lagging (24.58°)
    • Regulation 15.74% (|A|=0.9754); copper loss 0.8144 MW → efficiency 93.65%; labelled phasor diagram (demanded) provided

    Loses marks

    • Formula application without equivalent circuit
    • Sign errors in phasor calculations
    • Missing units in final results

    Earns more

    • Draw accurate phasor diagram
    • Show step-by-step complex number calculations
    • State all assumptions clearly
    • Include units in all intermediate steps

    Extra mark

    • Provide per-unit system analysis
    • Comment on line stability or power flow
  2. (b) Find pulse spacing for 8 kHz sampling and Nyquist rate sampling. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • 25 pulses per frame (24 + sync); τ=1 µs
    • 8 kHz: Ts=125 µs → slot 5 µs; clear spacing 4 µs (leading-edge 5 µs)
    • Nyquist 6.8 kHz: Ts=147.06 µs → slot 5.882 µs; clear spacing 4.882 µs
    • Correct frame arithmetic for both cases

    Loses marks

    • Forgetting synchronization pulse in channel count
    • Using wrong Nyquist rate formula
    • Missing units in time calculations

    Earns more

    • Draw timing diagram of multiplexed signal
    • Show flat-top sampling pulse duration
    • State Nyquist theorem explicitly
    • Include synchronization pulse in calculations

    Extra mark

    • Discuss guard time requirements
    • Compare bandwidth efficiency of both cases
  3. (c) Determine earthing resistance for 90% protection and protected percentage for 15Ω resistance. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • CT ratio 20 → pickup primary current 20 A
    • Re for 90% protection: Re=0.10×19052.56/20≈95.26 Ω
    • Re=15 Ω: α=20×15/19052.56=0.01575 → 1.57% unprotected
    • Protected winding ≈98.43%

    Loses marks

    • Wrong CT ratio application
    • Missing star connection voltage calculation
    • Incorrect fault current calculation

    Earns more

    • Show CT ratio conversion clearly
    • Draw phasor diagram for fault condition
    • State assumptions about fault location
    • Include all unit conversions

    Extra mark

    • Discuss sensitivity of protection scheme
    • Comment on relay operating characteristics
What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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