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Electrical Engineering 2021 Paper I 50 marks Solve

Paper I — Q6

(a) The magnetic field intensity of a linearly polarized uniform plane wave propagating in the +Y-direction in sea water (ε_r =…

(a)

The magnetic field intensity of a linearly polarized uniform plane wave propagating in the +Y-direction in sea water (ε_r = 80, μ_r = 1, σ = 4 S/m) is H = 0·1 sin (10^10 πt - π/3) a_x A/m. At Y = 0, determine the following: (i) The attenuation constant, intrinsic impedance, the wavelength and skin depth. (ii) The location at which the amplitude of H is 0·01 A/m. (iii) The expression for E(y, t) and H(y, t) at Y = 0·5 (m) as functions of t. 20 marks

(b)

A three-phase bridge inverter shown in Figure 6(b) is used to feed a Y-connected resistive load with R = 10 Ω per phase. The dc input to the inverter V_S = 400 V and the output frequency is 50 Hz. If the inverter is operating with 180° conduction mode, (i) compute the rms value of the load current, (ii) compute the rms value of the current in each switching device, (iii) calculate the output power, and (iv) draw the waveforms of phase and line voltages. 20 marks

(c)

Let the measurement error of a physical quantity be defined by a random variable X and its density function as follows: f(x) = {K(3-x²) for -1≤x≤1, {0 elsewhere. Determine the value of 'K' and find the probability that a random error in measurement is less than 1/2. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

समुद्र के जल (ε_r = 80, μ_r = 1, σ = 4 S/m) में +Y-दिशा में संचरित एक रेखीय ध्रुवित एकसमान समतल तरंग के चुंबकीय क्षेत्र की तीव्रता H = 0·1 sin (10^10 πt - π/3) a_x A/m है । Y = 0 पर निम्न का मान ज्ञात कीजिए : (i) शीणन स्थिरांक, नैज (इंट्रिंसिक) प्रतिबाधा, तरंगदैर्ध्य तथा सतही गहराई (स्किन डेप्थ) । (ii) वह स्थान जहाँ H का आयाम 0·01 A/m है । (iii) Y = 0·5 (m) पर t के फलन के रूप में E(y, t) तथा H(y, t) के लिए व्यंजक । (20 अंक)

(b)

चित्र 6(b) में प्रदर्शित एक त्रि-कला सेतु (ब्रिज) प्रतिलोमक (इन्वर्टर), एक Y-संयोजित R = 10 Ω प्रति कला वाले प्रतिरोधीय भार को भारित करने के लिए प्रयुक्त किया जाता है । प्रतिलोमक में dc निवेश V_S = 400 V तथा निगत आवृत्ति 50 Hz है । यदि प्रतिलोमक 180° चालकता विधा में कार्यशील है, तो (i) भार धारा का rms मान ज्ञात कीजिए, (ii) प्रत्येक स्विचिंग युक्ति में धारा का rms मान ज्ञात कीजिए, (iii) निर्गत शक्ति ज्ञात कीजिए, और (iv) कला वोल्टता तथा लाइन वोल्टता के तरंग रूपों को आरेखित कीजिए । (20 अंक)

(c)

मान लीजिए कि एक भौतिक राशि की मापन त्रुटि एक यादृच्छिक चर X तथा इसका घनत्व फलन निम्नलिखित है : f(x) = {K(3-x²) for -1≤x≤1, {0 अन्यथा । तो, 'K' के मान की गणना कीजिए तथा मापन में यादृच्छिक त्रुटि के 1/2 से कम होने की प्रायिकता ज्ञात कीजिए । (10 अंक)

Q6 of the 2021 UPSC Mains Electrical Engineering Paper I, as printed
The question as printed in the 2021 Electrical Engineering paper
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The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(b) Figure 6(b) shows a three-phase bridge inverter circuit. A DC voltage source Vs is connected to the top and bottom DC bus rails. The circuit consists of three vertical legs, each containing two NPN transistors (Q1, Q3, Q5 in the top half; Q4, Q6, Q2 in the bottom half) connected in series between the rails. Each transistor has a diode connected in anti-parallel (cathode to collector, anode to emitter). The junction between the top and bottom transistors of each leg forms a phase output terminal: node 'a' (between Q1 and Q4), node 'b' (between Q3 and Q6), and node 'c' (between Q5 and Q2). These three nodes connect to a Y-connected resistive load. The three resistors, each labeled R, are connected from nodes a, b, and c to a common neutral point labeled 'n'.

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) Plane wave in sea water ω = π × 10¹⁰ rad/s (5 GHz), ε = 80ε₀ with ε₀ = 8.854 × 10⁻¹² F/m, μ = μ₀. Loss tangent σ/(ωε) = 4/(π × 10¹⁰ × 80 × 8.854 × 10⁻¹²) = 0.18, so sea water is a lossy dielectric at this frequency and the exact formulas are used. (i) γ = √[jωμ(σ + jωε)] = α + jβ: α = 83.9 Np/m, β = 941.0 rad/m. η = √[jωμ/(σ + jωε)] = 41.8∠5.1° Ω. λ = 2π/β = 6.68 mm; skin depth δ = 1/α = 11.9 mm. (ii) |H| = 0.1e^(−αy) = 0.01 gives y = ln 10/α = 27.4 mm. (iii) E × H must point along +y; with H along a_x, E is along a_z, with amplitude |η|H₀ and leading H by 5.1°: H(y, t) = 0.1e^(−83.9y) sin(10¹⁰πt − 941.0y − 60°) a_x A/m E(y, t) = 4.18e^(−83.9y) sin(10¹⁰πt − 941.0y − 60° + 5.1°) a_z V/m At y = 0.5 m, e^(−αy) = 6.03 × 10⁻¹⁹ and βy = 470.5 rad ≡ 318.5°, so H = 6.0 × 10⁻²⁰ sin(10¹⁰πt − 18.5°) a_x A/m E = 2.5 × 10⁻¹⁸ sin(10¹⁰πt − 13.4°) a_z V/m (The reduced phase is very sensitive to rounding of β, since βy covers about 75 cycles.) Half a metre is about 42 skin depths, so the field has all but vanished.

(b) Three-phase bridge inverter, 180° conduction Three devices conduct at any instant, so each phase sees V_s/3 or 2V_s/3 and the phase voltage is a six-step wave: V_s/3, 2V_s/3, V_s/3, −V_s/3, −2V_s/3, −V_s/3 in 60° steps. (i) V_ph(rms) = √[(1/3)((V_s/3)² + (2V_s/3)² + (V_s/3)²)] = (√2/3)V_s = 188.6 V, so I_L(rms) = 18.86 A. (ii) The load is resistive, so the diodes never conduct. Each transistor carries the phase current for one half-cycle: I_Q(rms) = I_L/√2 = 13.33 A (average 8.89 A). (iii) P = 3I_L²R = 3 × 18.86² × 10 = 10.67 kW. (iv) Phase voltage: the six-step wave above, with v_bn and v_cn shifted by 120° and 240°. Line voltage: a quasi-square wave of height ±V_s = ±400 V, 120° wide, with 60° zeros between, V_L(rms) = √(2/3)V_s = 326.6 V; v_bc and v_ca are v_ab shifted by 120° and 240°.

Diagram: 180-degree three-phase inverter: phase and line voltages

(c) Density constant and probability ∫₋₁¹ K(3 − x²)dx = K[3x − x³/3]₋₁¹ = K(16/3) = 1, so K = 3/16. P(X < ½) = (3/16)[3x − x³/3]₋₁^(1/2) = (3/16)[(3/2 − 1/24) + 8/3] = (3/16)(99/24) = 99/128 ≈ 0.773. (If "error less than ½" is read as a magnitude, P(|X| < ½) = 35/64 ≈ 0.547.)

How this answer will be evaluated

What earns full marks: All calculations correct with proper units, complete waveforms, and clear probability derivation

Evaluation rubric

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

  1. (a) Wave parameters, location of a given amplitude and field expressions for a plane wave in sea water. 20 marks

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

    Must cover

    • Check the loss tangent (0.18) to choose the exact formulas, and obtain α = 83.9 Np/m, β = 941.0 rad/m, η = 41.8∠5.1° Ω, λ = 6.68 mm and δ = 11.9 mm.
    • Find y = ln 10/α = 27.4 mm where |H| = 0.01 A/m.
    • Write H and E at y = 0.5 m with the correct directions (E along a_z) and phases (about −18.5° and −13.4°).

    Loses marks

    • Using the good-conductor approximation without checking σ/(ωε).
    • Wrong sign of the phase term or wrong direction of E.

    Earns more

    • Use of the complex propagation constant γ.
    • E leading H by the angle of η and the amplitude ratio |η|.

    Extra mark

    • Remark that 0.5 m is about 42 skin depths, so the field has all but vanished.
  2. (b) RMS load current, RMS device current, output power and voltage waveforms of a three-phase bridge inverter in 180° conduction. 20 marks

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

    Must cover

    • Identify the six-step phase voltage (V_s/3, 2V_s/3, V_s/3, …) and get V_ph(rms) = 188.6 V and I_L = 18.86 A.
    • Find I_Q(rms) = 13.33 A (each transistor conducts half a cycle; the diodes do not conduct for a resistive load) and P = 10.67 kW.
    • Draw the six-step phase voltage and the 120° quasi-square line voltage with the levels marked.

    Loses marks

    • Taking the phase voltage as a square wave of V_s/2.
    • Wrong rms method or missing waveforms.

    Earns more

    • Conduction sequence of the devices stated.
    • Waveforms labelled with voltage levels and 120° shifts.

    Extra mark

    • Line-voltage rms 326.6 V = √3 × V_ph.
  3. (c) Normalising constant K and the probability that the error is less than ½. 10 marks

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

    Must cover

    • Integrate f(x) over −1 to 1 to get K = 3/16.
    • Integrate over −1 to ½ to get P(X < ½) = 99/128.
    • State the result as about 0.773.

    Loses marks

    • Wrong normalisation such as K = 3/4.
    • Wrong lower limit or arithmetic slips in the integral.

    Earns more

    • Correct limits and antiderivative 3x − x³/3.
    • Clear final statement of the probability.

    Extra mark

    • Alternative reading P(|X| < ½) = 35/64.
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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