Electrical Engineering 2025 Paper I 50 marks Compulsory Solve

Q5

(a) As shown in the figure, just inside the surface of a dielectric slab, the electric field (E₁) is 15 V/m and it makes an angle of 30° with the surface. The electric field (E₂) makes 65.5° angle with the surface, just above the surface. Determine the magnitude of E₂ and the dielectric constant of the slab. (10 marks) (b) Show with the help of suitable derivations that the voltage regulation of a transformer varies with the power factor of the load. At what power factor will the voltage regulation be : (i) zero, and (ii) maximum ? (10 marks) (c) A single-phase Thyristor converter circuit as shown in the figure is feeding to a constant current load of 10 A. The supply voltage is of 230 V, 50 Hz and source inductance of 2 mH. Assume the Thyristors are ideal and triggering angle α = 30°. Calculate (i) the overlap angle u, and (ii) the drop in output voltage. (10 marks) (d) Show that for a binomial random variable, the mean is given by np and the variance is given by np (1 – p), where n gives the number of trials and p gives the probability of successes. (10 marks) (e) The frequency range of operation of a superheterodyne FM receiver is 88 MHz – 108 MHz. The centre frequency of the IF amplifier (f_IF) and the frequency of the local oscillator (f_LO) are so chosen that f_IF < f_LO. The design has to be so carried out that the image frequency f'_c falls outside of the 88 MHz – 108 MHz region. Determine the minimum required value of f_IF and the corresponding range of variations in f_LO for that chosen value of f_IF. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) जैसा कि यहाँ चित्र में प्रदर्शित है, एक परावैद्युत गुटके (स्लैब) की सतह के ठीक अंदर विद्युत-क्षेत्र E₁ = 15 V/m है और यह सतह से 30° का कोण बनाता है। सतह के ठीक ऊपर विद्युत-क्षेत्र (E₂) सतह से 65.5° का कोण बनाता है। E₂ का परिमाण तथा गुटके (स्लैब) का परावैद्युतांक निर्धारित कीजिए। (10 अंक) (b) उपयुक्त न्युतियों की सहायता से दर्शाइए कि एक परिणामित्र का वोल्टता नियमन भार के शक्ति गुणक के साथ बदलता है। किस शक्ति गुणक पर वोल्टता नियमन, (i) शून्य होगा, एवं (ii) अधिकतम होगा? (10 अंक) (c) चित्र में दर्शाए गए अनुसार एक, एककलीय थाइरिस्टर परिवर्तित्र परिपथ एक 10 A के स्थिर धारा भार को पूरित करता है। प्रदाय वोल्टता 230 V, 50 Hz और स्रोत प्रेरकत्व 2 mH है। मान लीजिए कि थाइरिस्टर आदर्श हैं तथा ट्रिगर कोण α = 30° है, तो (i) अतिव्यापन कोण u, और (ii) निर्गत वोल्टता में अवपातन की गणना कीजिए। (10 अंक) (d) दर्शाइए कि एक द्विपद यादृच्छिक चर के लिए माध्य np द्वारा तथा प्रसरण np (1 – p) द्वारा निर्धारित होता है, जहाँ n परीक्षणों की संख्या तथा p सफलताओं की संभावना (प्रायिकता) प्रदर्शित कर रहे हैं। (10 अंक) (e) एक सुपरहेटेरोडाइन FM अभिग्राहित की कार्यशीलता का आवृत्ति परास 88 MHz – 108 MHz है। IF प्रवर्धक की मध्य आवृत्ति (f_IF) तथा स्थानीय दोलित्र की आवृत्ति (f_LO) इस प्रकार चयनित हैं कि f_IF < f_LO है। अभिकल्पना इस प्रकार की जानी है ताकि प्रतिबिंब आवृत्ति f'_c, 88 MHz – 108 MHz क्षेत्र के बाहर हो। f_IF का न्यूनतम आवश्यक मान और उस चुने गए f_IF के मान के लिए f_LO के संबंधित बदलाव का परास निर्धारित कीजिए। (10 अंक)

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Approach

This is a multi-part problem-solving question requiring equal attention to all five sub-parts (10 marks each). Begin with a brief introduction acknowledging the diverse topics covered. For part (a), apply boundary conditions for electric fields at dielectric interfaces. For part (b), derive the voltage regulation expression using transformer equivalent circuit. For part (c), analyze the single-phase converter with source inductance considering overlap. For part (d), prove mean and variance using binomial distribution properties. For part (e), apply superheterodyne receiver principles to determine IF and LO ranges. Allocate approximately equal time (~18-20 minutes) per sub-part, presenting each solution clearly with proper headings.

Key points expected

  • Part (a): Apply tangential E continuity (E₁sinθ₁ = E₂sinθ₂) and normal D continuity (ε₁E₁cosθ₁ = ε₂E₂cosθ₂) to find εᵣ = 3.0 and E₂ = 8.66 V/m
  • Part (b): Derive voltage regulation as %R = (I₂R₀₂cosφ ± I₂X₀₂sinφ)/V₂ × 100, showing zero regulation at leading pf = R₀₂/√(R₀₂²+X₀₂²) and maximum at lagging unity pf
  • Part (c): Calculate overlap angle u = 4.2° using cos(α+u) = cosα - (2ωLₛI₀)/Vₘ, and voltage drop ΔV₀ = (ωLₛI₀/π) = 2 V
  • Part (d): Prove E[X] = np using Σk·ⁿCₖpᵏqⁿ⁻ᵏ = np(p+q)ⁿ⁻¹ = np, and Var(X) = np(1-p) using E[X²] - (E[X])²
  • Part (e): Determine f_IF(min) = 10 MHz ensuring image frequency f'c = f_LO + f_IF = fc + 2f_IF > 108 MHz, giving LO range 98-118 MHz

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies and applies: boundary conditions for electrostatic fields in (a), transformer equivalent circuit and phasor relationships in (b), converter commutation with source inductance in (c), binomial probability mass function properties in (d), and superheterodyne image frequency rejection in (e); no conceptual errors across any sub-partMostly correct concepts with minor errors in 1-2 sub-parts, such as confusing boundary conditions or misapplying voltage regulation sign conventions; understands core principles but with gapsMajor conceptual errors in multiple sub-parts, such as applying Snell's law incorrectly in (a), omitting reactive components in (b), ignoring overlap in (c), or fundamental misunderstanding of binomial distribution in (d)
Numerical accuracy20%10All calculations precise: (a) εᵣ = 3.0, E₂ = 8.66 V/m; (c) u ≈ 4.2°, voltage drop = 2 V; (e) f_IF = 10 MHz with correct LO range; proper significant figures and unit handling throughoutCorrect methodology but arithmetic errors in 1-2 sub-parts, or correct final answers with intermediate calculation omissions; minor unit conversion errorsMultiple calculation errors, wrong formulas leading to incorrect magnitudes, missing units, or order-of-magnitude mistakes indicating poor numerical handling
Diagram quality20%10Clear labeled diagrams for (a) showing field vectors and angles at dielectric interface, (b) transformer equivalent circuit with phasor diagram, (c) converter waveforms with overlap period marked, and (e) frequency spectrum showing RF, LO, IF and image positions; neat, properly dimensionedDiagrams present but lacking labels or clarity in 1-2 parts; missing phasor diagrams in (b) or waveform sketches in (c); roughly drawn but recognizableMissing critical diagrams, unrecognizable sketches, or diagrams that contradict the written solution; no attempt at visual representation for parts clearly needing figures
Step-by-step derivation20%10Complete logical flow: (a) explicit statement of two boundary conditions before solving, (b) full derivation from equivalent circuit to regulation formula with condition analysis, (c) integration of commutation equation for overlap, (d) rigorous summation derivation with factorial manipulation, (e) systematic inequality setup for image rejectionCorrect final formulas but skips key derivation steps; jumps from given data to answer without showing boundary condition application or summation steps; logical gaps presentNo derivations shown, only final answers stated; or incorrect derivations with algebraic errors; missing essential steps that prevent verification of understanding
Practical interpretation20%10Connects theory to practice: (a) mentions capacitor dielectric applications, (b) discusses transformer loading strategies for Indian grid conditions, (c) notes thyristor protection during commutation, (d) relates to quality control sampling, (e) explains why 10.7 MHz standard IF used in FM receivers; insightful engineering judgmentBrief mention of practical relevance in 2-3 parts without elaboration; generic statements about importance without specific application contextPurely mathematical treatment with no physical interpretation; fails to explain what results mean for device operation or system design; no real-world context provided

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