Electrical Engineering 2023 Paper I 50 marks Compulsory Solve

Q1

(a) Obtain Norton equivalent circuit at terminals ab of the coupled circuit shown in the figure. Using it, find out the current passing through 5 Ω resistor connected between the terminals ab. (10 marks) (b) Obtain the Laplace transform of the following periodic waveforms : (10 marks) (i) (ii) (c) A 3-phase, 50 Hz, star-connected cage-type induction motor has standstill input impedance of (1·0 + j 3·0) Ω per phase. The motor is connected through a cable from 400 V, 3-phase balanced supply so that the blocked rotor voltage at its terminal is dropped by 20% from the supplied voltage. The motor is to be started through a DOL starter from the same supply and cable as above. Find : (i) the cable impedance per phase, (ii) the motor starting current, (iii) input power factor at the time of starting. (Assume negligible stator impedance of the motor and cable R/X ratio of 3 : 1 at 50 Hz supply. Also ignore magnetizing current and core losses.) (10 marks) (d) Calculate the lower corner frequency for the circuit shown below. Take transistor parameters as : β = 100, V_BE = 0·7 V and V_A = ∞. (10 marks) V_CC = 12 V, R_1 = 10 kΩ, R_S = 0·5 kΩ, C_C = 0·1 μF, R_2 = 1·5 kΩ, R_C = 1 kΩ, R_E = 0·1 kΩ (e) A metal bar slides over a pair of conducting rails in a uniform magnetic field B⃗ = a⃗_z B_0 Wb/m² with a constant velocity u⃗ m/s as shown below in the figure. A resistance 'R' Ω is connected between terminals 1 and 2. Prove that this system upholds the principle of conservation of energy. Neglect the electrical resistance of the metal bar and the pair of conducting rails, and the mechanical friction of this ideal system. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) चित्र में दर्शाए गए युग्मित परिपथ का ab अन्तस्थों पर नॉर्टन समतुल्य परिपथ निकालिए । उसका उपयोग करते हुए, ab अन्तस्थों के मध्य जुड़े 5 Ω के प्रतिरोधक से प्रवाहित होने वाली धारा ज्ञात कीजिए । (10 अंक) (b) निम्नलिखित आवर्ती तरंगरूपों का लाप्लास रूपांतरण निकालिए : (10 अंक) (i) (ii) (c) एक त्रि-कला, 50 Hz तारा-संयोजित पंजर प्रेरण मोटर की विराम निवेश प्रतिबाधा (1·0 + j 3·0) Ω प्रति कला है । यह मोटर 400 V, त्रि-कला संतुलित विद्युत प्रदाय से केबल के माध्यम से जोड़ी गई है जिससे इसके सिरे पर अवरोधित घूर्णक बोल्टता संभरण बोल्टता से 20% कम हो । इस मोटर को उपयुक्त संभरण बोल्टता एवं केबल द्वारा एक DOL स्टार्टर के माध्यम से शुरु किया जाना है । ज्ञात कीजिए : (i) केबल की प्रतिबाधा प्रति कला, (ii) मोटर की आरंभिक धारा, (iii) मोटर के शुरू होते समय निवेश शक्ति गुणांक । (मोटर की स्टेटर प्रतिबाधा को नगण्य एवं 50 Hz विद्युत प्रदाय पर केबल का R/X अनुपात 3 : 1 मानिए । साथ ही चुंबकीय धारा एवं क्रोड हानियों को भी अनदेखा कीजिए ।) (10 अंक) (d) नीचे दर्शाए गए परिपथ के लिए निम्न कोणा आवृत्ति निकालिए । ट्रांजिस्टर के प्राचल निम्नानुसार हैं : β = 100, V_BE = 0·7 V एवं V_A = ∞ । (10 अंक) (e) चित्र में दर्शाए अनुसार एकसमान चुंबकीय क्षेत्र B⃗ = a⃗_z B_0 Wb/m² में एक धातु पट्टी एक सुचालक रेल के जोड़े के ऊपर अचर बेग u⃗ m/s से फिसलती है । अंतस्थ 1 और 2 के मध्य एक प्रतिरोध 'R' Ω जुड़ा है । सिद्ध कीजिए कि यह तंत्र ऊर्जा संरक्षण के सिद्धांत का अनुमोदन करता है । धातु पट्टी एवं सुचालक रेल के जोड़े के विद्युतीय प्रतिरोध तथा इस आदर्श तंत्र के यांत्रिक घर्षण को अनदेखा कीजिए । (10 अंक)

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Approach

Solve each sub-part systematically, allocating approximately 20% time to each 10-mark section. Begin with clear circuit diagrams for parts (a), (d), and (e), then apply standard network theorems, Laplace transform techniques, machine equations, and electromagnetic principles. Present derivations stepwise with final boxed answers for numerical quantities.

Key points expected

  • Part (a): Correct application of Norton's theorem to coupled circuits with proper handling of mutual inductance; calculation of short-circuit current and equivalent impedance; final current through 5Ω resistor
  • Part (b): Application of periodic Laplace transform formula F(s) = (1/(1-e^(-sT)))∫[0 to T]f(t)e^(-st)dt for both waveforms; correct identification of period and piecewise functions
  • Part (c): Calculation of cable impedance from 20% voltage drop condition; blocked rotor current using standstill impedance; starting current and power factor with cable impedance in series
  • Part (d): DC analysis for operating point; small-signal model for CE amplifier; calculation of lower corner frequency f_L = 1/(2π(R_S + r_π)C_C) with proper input resistance reflection
  • Part (e): Derivation of motional EMF (ε = B₀lu); power delivered to resistance (P = ε²/R = B₀²l²u²/R); mechanical power input (F = B₀²l²u/R, P_mech = Fu); equality proof for energy conservation

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies Norton's theorem with mutual inductance, uses periodic Laplace transform formula properly, applies 3-phase induction motor equivalent circuit with cable drop, sets up small-signal BJT model correctly, and derives motional EMF with force balance for energy conservationMinor errors in theorem application such as sign convention in mutual inductance, incorrect period in Laplace integral, or missing cable impedance in motor circuit; basic understanding evident but some conceptual gapsFundamental misconceptions like treating coupled inductors as independent, using unilateral Laplace for periodic functions without modification, or confusing blocked rotor with running conditions; energy conservation proof missing key elements
Numerical accuracy20%10All numerical values accurate: Norton current and impedance, Laplace transforms in simplified form, cable impedance (0.2 + j0.067Ω), starting current (~92A at 0.316 pf), corner frequency (~3.18 Hz), and exact power balance in (e)Correct methodology with arithmetic errors or unit conversion mistakes; answers in correct order of magnitude but final values slightly off; partial credit for correct setupGross calculation errors, wrong formulas leading to absurd values, or missing numerical answers entirely; no demonstration of quantitative reasoning
Diagram quality15%7.5Clear original circuit diagrams for (a), (d), (e) with all components labeled; Norton equivalent circuit drawn; small-signal equivalent for BJT amplifier; rail-bar system with velocity and field directions; proper use of standard symbolsDiagrams present but missing labels or with incorrect connections; hand-drawn appearance acceptable but clarity compromised; essential elements present but not professionally presentedMissing diagrams where required, or diagrams that misrepresent the physical system; no attempt to illustrate circuit configurations or electromagnetic setup
Step-by-step derivation25%12.5Systematic progression from given data to solution: mesh/nodal analysis for Norton, integral setup and evaluation for Laplace, voltage divider and impedance combination for motor, DC bias then small-signal for amplifier, EMF derivation then power balance for energy proof; each step justifiedSome steps skipped or combined without justification; final answer correct but derivation unclear; missing intermediate results that would aid verificationNo logical flow, equations presented without connection to problem, or direct assertion of answers without derivation; impossible to follow reasoning
Practical interpretation20%10Interprets results practically: discusses significance of Norton equivalent for load analysis, relevance of Laplace for system response, implications of high starting current and low pf for DOL starter design, frequency response for amplifier coupling, and validates energy conservation as fundamental principle for electromechanical systemsBrief mention of practical relevance without elaboration; standard concluding statements without specific connection to calculated valuesNo interpretation of results; purely mathematical exercise without engineering context; misses opportunity to relate to real-world applications like motor protection or amplifier design

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