Electrical Engineering 2023 Paper II 50 marks Solve

Q8

(a) The figure shows a unity feedback system with G(s) = 2/s(s+1)(s+2)(s+3). (i) Sketch the approximate polar plot of G(s). (ii) Determine the closed-loop stability of the system using the polar plot of G(s). (iii) Determine the gain margin of the system. 20 (b) Two ammeters having resistances of 1 Ω and 2 Ω respectively give full-scale deflections with 200 mA and 250 mA respectively. Find the shunt to be connected with these ammeters to extend their range to 20 A. The range extended ammeters are connected in parallel and then placed in a circuit in which a total current of 15 A is flowing. Find the readings of the ammeters. 20 (c) Two generating stations having short-circuit capacities of 1500 MVA and 1000 MVA respectively, and operating at 11 kV, are linked by an interconnected cable having a reactance of 0·7 Ω per phase. Determine the short-circuit capacity of each station after interconnection. 10

हिंदी में प्रश्न पढ़ें

(a) चित्र में G(s) = 2/s(s+1)(s+2)(s+3) के साथ एक इकाई पुनर्निवेश तंत्र प्रदर्शित की गई है। (i) G(s) का सरलिकृत ध्रुवी आरेख बनाइए। (ii) G(s) के ध्रुवी आरेख को उपयोग में लेते हुए तंत्र का बंद-लूप स्थायित्व ज्ञात कीजिए। (iii) तंत्र का लब्धि उपांत (गेन मार्जिन) ज्ञात कीजिए। 20 (b) दो ऐमीटर, जिनके प्रतिरोध क्रमशः: 1 Ω एवं 2 Ω हैं, के पूर्ण-पैमाना विचेष क्रमशः: 200 mA एवं 250 mA हैं। इन ऐमीटरों के पारस को 20 A तक बढ़ाने के लिए संयोजित किया जाने वाला शंट ज्ञात कीजिए। पारस वर्धित ऐमीटरों (रेंज एक्सटेंडेड ऐमीटरों) को समांतर क्रम में जोड़ा गया है और फिर उनको एक ऐसे परिपथ में रखा गया है जिसमें कुल धारा 15 A बह रही है। दोनों ऐमीटरों के पाठ्यांकों (रीडिंग्स) को ज्ञात कीजिए। 20 (c) दो विद्युत-उत्पादन केन्द्र, जिनकी लघुपथ क्षमताएँ क्रमशः: 1500 MVA एवं 1000 MVA हैं और 11 kV पर प्रचालित हैं, 0·7 Ω प्रतिघात प्रति कला वाली एक केबल द्वारा अन्तःसंयोजित (इन्टरकनेक्टेड) हैं। अन्तःसंयोजन के उपरांत प्रत्येक विद्युत-उत्पादन केन्द्र की लघुपथ क्षमता ज्ञात कीजिए। 10

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How this answer will be evaluated

Approach

Solve all three parts systematically, allocating approximately 40% time to part (a) given its 20 marks and complexity involving polar plots and stability analysis; 35% to part (b) for shunt calculations and current distribution; and 25% to part (c) for fault level calculations. Begin with clear problem statements, show all derivations with proper units, and conclude with physical interpretations of results.

Key points expected

  • For (a)(i): Calculate magnitude and phase of G(jω) at key frequencies (ω→0, ω→∞, and phase crossover), identify type number (n=1) and starting angle (-90°), sketch polar plot showing encirclement pattern
  • For (a)(ii): Apply Nyquist stability criterion correctly, count encirclements of (-1+j0) point, determine N and P, conclude closed-loop stability with Z = P - N = 0
  • For (a)(iii): Calculate gain margin as 1/|G(jω_pc)| where phase crossover frequency ω_pc satisfies ∠G(jω_pc) = -180°, express in dB
  • For (b): Calculate shunt resistances R_sh1 = (1×0.2)/(20-0.2) and R_sh2 = (2×0.25)/(20-0.25), then find equivalent resistances of extended-range ammeters, solve current divider for 15A total current
  • For (c): Convert short-circuit capacities to equivalent reactances (X1 = 11²/1500, X2 = 11²/1000), combine with interconnector reactance (0.7Ω), calculate new fault levels at each bus using Thevenin equivalent
  • For (c): Recognize that interconnected system modifies fault levels due to mutual contribution through tie-line, calculate S_sc_new at each station considering parallel contribution from other station

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies Nyquist criterion for (a)(ii) with proper identification of P=0 (no RHP poles in open-loop); accurately uses shunt formula m/(I-I_m) for range extension in (b); properly converts MVA fault levels to per-unit reactances and accounts for interconnector effect in (c); no conceptual errors in any sub-partMinor errors in applying Nyquist criterion (e.g., confusion about encirclement direction) or in fault level calculation (e.g., ignoring interconnector reactance in series); shunt calculations essentially correct but with unit conversion slipsFundamental misunderstanding of stability criterion (confusing open-loop and closed-loop poles), incorrect shunt formula application, or treating fault levels as simple addition without network analysis
Numerical accuracy20%10Precise calculations: gain margin within ±0.5 dB or exact fraction; shunt resistances R_sh1 ≈ 0.0101Ω, R_sh2 ≈ 0.0253Ω; ammeter readings sum correctly to 15A; fault levels correctly showing increase at weaker station and decrease at stronger station; all unit conversions (kV, MVA, Ω) handled flawlesslyCorrect method but arithmetic errors leading to 5-10% deviation in final answers; minor errors in current divider calculation or per-unit conversion; final answers plausible but not exactMajor calculation errors (order of magnitude mistakes), incorrect current distribution (readings not summing to 15A), or fault levels calculated without proper reactance combination
Diagram quality15%7.5Clear polar plot for (a)(i) with properly labeled axes, indication of (-1+j0) point, direction arrows showing ω increasing, asymptotic behavior at ω→0 and ω→0; neat circuit diagrams for (b) showing shunt connections and parallel arrangement; single-line diagram for (c) showing two stations and interconnector with reactance valuesPolar plot sketch roughly correct but missing key features (no (-1+j0) marking, unclear scale); minimal or no diagrams for parts (b) and (c); diagrams present but poorly labeledNo polar plot or completely incorrect sketch (e.g., wrong quadrant, missing 270° phase at high frequency); absence of circuit diagrams where essential for understanding
Step-by-step derivation25%12.5Complete stepwise working: for (a) explicit G(jω) substitution, magnitude and phase expressions, phase crossover equation solution; for (b) clear derivation of shunt formula from I_mR_m = (I-I_m)R_sh, verification of extended range; for (c) systematic conversion to reactances, Thevenin equivalent construction, fault MVA calculation with formula S_sc = V²/X_thSome steps skipped or combined (e.g., jumping to final formula without derivation), but logical flow maintained; minor gaps in algebraic manipulation; final answers obtained but process not fully transparentAnswers stated without derivation, missing critical steps (e.g., no phase crossover calculation, direct assertion of gain margin), or disorganized presentation making verification impossible
Practical interpretation20%10Interprets (a) gain margin in terms of system robustness and permissible gain increase before instability; explains (b) ammeter loading effect and why readings differ despite same nominal range; discusses (c) interconnection benefits (mutual support during faults) and how weaker station's fault level improves while stronger station's fault level decreases, relevant to Indian grid interconnection practicesBrief mention of practical significance without elaboration; generic statements about stability or measurement accuracy; minimal discussion of grid interconnection benefitsPurely mathematical treatment with no physical interpretation; no discussion of why results matter in engineering practice; missing insight into interconnected system behavior

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