Electrical Engineering 2023 Paper II 50 marks Compulsory Solve

Q1

(a) The figure shows a unity feedback system. The steady-state value of the unit step response c(t) is 0·8. Determine the maximum overshoot in the response c(t) : (10 marks) (b) A circuit breaker is rated as 2500 A, 1500 MVA, 33 kV, 3 sec, 3-phase, oil circuit breaker. Determine its rated normal current, breaking current, making current and short-time rating (current). (10 marks) (c) An audio signal, whose bandwidth is 15 kHz, is to be digitized using PCM. Uniform quantization with 1024 levels and binary encoding are assumed. Determine the minimum sampling rate. If the actual sampling rate is 20% excess of the minimum rate, determine the minimum permissible bit rate. (10 marks) (d) Briefly explain the following logical instructions of 8085 microprocessor : (i) ANA M (ii) XRA M (iii) CMC (iv) STC (v) RRC (10 marks) (e) In a three-phase 400 km long transmission line, the conductors are spaced at the corners of an equilateral triangle of side 5 m. The diameter of each conductor is 3 cm. Calculate the capacitance per phase of the 400 km long conductor. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) चित्र में एक इकाई पुनर्निवेश तंत्र दिखाया गया है। इकाई पद अनुक्रिया (रेस्पोंस) c(t) का स्थायी-दशा मान 0·8 है। अनुक्रिया c(t) में अधिकतम ओवरशूट ज्ञात कीजिए : (10 अंक) (b) एक परिपथ विचोजक को 2500 A, 1500 MVA, 33 kV, 3 sec, 3-कला, ऑयल परिपथ विचोजक की तरह निर्धारित किया गया है। उसकी निर्धारित सामान्य धारा, विचोजन धारा, संयोजन धारा एवं लघु-समय रेटिंग (धारा) ज्ञात कीजिए। (10 अंक) (c) एक श्रव्य संकेत, जिसकी बैंड-चौड़ाई 15 kHz है, को पी० सी० एम० का उपयोग करते हुए अंकिक बनाना है। 1024 स्तरों (levels) के साथ एकसमान कांटन एवं द्वि-आधारी कोडन मान लीजिए। न्यूनतम प्रतिचयन दर (मिनिमम सैम्पलिंग रेट) निर्धारित कीजिए। यदि वास्तविक प्रतिचयन दर, न्यूनतम दर से 20% ज्यादा है, तो न्यूनतम अनुमत्य बिट दर ज्ञात कीजिए। (10 अंक) (d) 8085 सूक्ष्म-संसाधित्र (माइक्रोप्रोसेसर) के निम्नलिखित तार्किक निर्देशों की संक्षिप्त व्याख्या कीजिए : (i) ANA M (ii) XRA M (iii) CMC (iv) STC (v) RRC (10 अंक) (e) एक त्रि-कला (श्री-फेज़) 400 km लम्बी संचरण लाइन में, चालकों (कण्डक्टरों) को 5 m भुजा वाले एक समबाहु त्रिभुज के कोनों के अन्तराल पर रखा गया है। प्रत्येक चालक का व्यास 3 cm है। इस 400 km लम्बे चालक की प्रति कला धारिता की गणना कीजिए। (10 अंक)

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Approach

Solve all five sub-parts systematically with equal time allocation (~20% each) since all carry equal marks. Begin with the control system problem (a) requiring steady-state error analysis and overshoot calculation, then proceed sequentially through circuit breaker ratings (b), PCM sampling and bit rate (c), 8085 instruction explanations (d), and transmission line capacitance (e). Present each solution with clear problem identification, relevant formulas, step-by-step working, and final boxed answers.

Key points expected

  • Part (a): Apply final value theorem to find K from steady-state value 0.8, then determine damping ratio ζ and natural frequency ωn, finally calculate percentage overshoot using Mp = exp(-πζ/√(1-ζ²)) × 100
  • Part (b): Identify rated normal current = 2500 A, breaking current = 2500 A (symmetrical), making current = 2.55 × breaking current (IEC standard), short-time rating = 2500 A for 3 seconds, MVA rating = √3 × 33 kV × breaking current
  • Part (c): Apply Nyquist criterion for minimum sampling rate = 2 × 15 kHz = 30 kHz, actual rate = 1.2 × 30 kHz = 36 kHz, bits per sample = log₂(1024) = 10, bit rate = 36 kHz × 10 = 360 kbps
  • Part (d): Explain ANA M (AND accumulator with memory), XRA M (XOR accumulator with memory), CMC (complement carry flag), STC (set carry flag), RRC (rotate accumulator right through carry) with flag effects
  • Part (e): Calculate capacitance per phase using C = 2πε₀/ln(D/r) per km, where D = 5 m, r = 1.5 cm, then multiply by 400 km; convert to μF or nF with proper units

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies unity feedback system type for (a), distinguishes all four circuit breaker ratings in (b), applies Nyquist-Shannon theorem precisely in (c), accurately describes all five 8085 instructions with flag effects in (d), and uses correct three-phase line capacitance formula with equivalent radius in (e)Minor conceptual errors in 1-2 parts such as confusing making current factor (using 1.8 instead of 2.55), or slight misapplication of sampling theorem, or incomplete flag descriptions for 8085 instructionsFundamental misconceptions like treating circuit breaker MVA as power consumption, applying incorrect overshoot formula, or confusing logical instructions with arithmetic instructions in 8085
Numerical accuracy20%10All calculations precise: overshoot ~2.84% for (a), making current ~6375 A for (b), bit rate exactly 360 kbps for (c), capacitance ~4.2 μF per phase for (e); proper unit handling throughoutCorrect method but arithmetic slips in 1-2 parts, such as calculation errors in logarithm for capacitance or rounding errors in overshoot percentage; units mostly correctMajor calculation errors in multiple parts, wrong order of magnitude in results, missing or inconsistent units, or failure to convert between kHz, Hz, km, m appropriately
Diagram quality15%7.5Clear block diagram for unity feedback system in (a) with R(s), C(s), G(s), H(s)=1 labeled; neat three-phase arrangement sketch for (e) showing equilateral triangle spacing; well-organized tabular presentation for 8085 instructions in (d)Basic diagrams present but lacking labels or clarity; rough sketches without proper notation; or reliance on text description where diagram would aid understandingNo diagrams where required, messy or misleading sketches, or complete omission of visual aids for parts (a) and (e) where spatial/geometric representation is essential
Step-by-step derivation25%12.5Complete derivation chain visible: from steady-state error formula to K value, then ζ and ωn extraction, finally overshoot formula in (a); explicit formula substitution with intermediate steps shown for capacitance in (e); systematic rating calculations in (b)Some steps skipped or condensed, 'therefore' statements without justification, or jumping from given data to final answer without showing key intermediate calculations in 2-3 partsNo derivation shown, only final answers stated, or incorrect formula application without explanation; missing essential steps like logarithm calculation for quantization levels in (c)
Practical interpretation20%10Relates 0.8 steady-state value to position error constant and Type 0 system in (a); explains significance of 3-sec rating for thermal withstand in (b); discusses 20% excess sampling for guard band against aliasing in practical PCM systems in (c); notes typical 8085 applications in industrial control for (d); comments on Ferranti effect implications of calculated capacitance in (e)Brief mention of practical relevance in 2-3 parts without elaboration, or generic statements not specific to the calculated valuesPurely mathematical treatment with no engineering context, failure to recognize why making current exceeds breaking current, or no appreciation of why 1024 levels implies 10-bit encoding

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