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Electrical Engineering 2023 Paper II 50 marks Compulsory Calculate

Paper II — Q5

(a) The figure shows a compensator network, where R₁ = 3 MΩ, R₂ = 1 MΩ, C = 1 μF. Vᵢ(t) and Vₒ(t) are the input voltage and…

(a)

The figure shows a compensator network, where R₁ = 3 MΩ, R₂ = 1 MΩ, C = 1 μF. Vᵢ(t) and Vₒ(t) are the input voltage and output voltage respectively. [Figure: R₁ in series with the input; the output is taken across C in series with R₂.] Determine the attenuation in dB provided by this network at very high frequencies : 10 marks

(b)

A resistive strain gauge, with a gauge factor 2·2, is cemented on a rectangular steel bar with the elastic modulus, E = 205×10⁶ kN/m². The width and thickness of the steel bar is 3·5 cm and 0·55 cm respectively. An axial force of 12 kN is applied. If the nominal resistance of the strain gauge is 100 Ω, determine the change in resistance of the strain gauge. 10 marks

(c)

A three-phase, 50 Hz, 415 V supply delivers 250 kW power at power factor of 0·8 lagging. The line power factor is desired to be improved to 0·9 lagging by installing shunt capacitors. Calculate the capacitance if they are connected in delta. 10 marks

(d)

Binary data is transmitted over additive white Gaussian noise (AWGN) channel at a bit rate of 5 kilobits/sec. The single-sided power spectral density for the channel is 10⁻⁷ W/Hz. Non-coherent orthogonal binary FSK with higher frequency signalling tone of 1 MHz is used. The bit energy, E_b = 2×10⁻⁶ J. Determine the minimum required bandwidth and average bit error probability. 10 marks

(e)

Consider a three-phase, Δ-Y connected, 30 MVA, 33/11 kV transformer with differential relay protection. If the CT ratios are 500 : 5A on the primary side and 2000 : 5A on the secondary side, compute the relay current setting for faults drawing up to 200% of rated transformer current. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

नीचे चित्र में एक प्रतिकारी संजाल प्रदर्शित है, जिसमें R₁ = 3 MΩ, R₂ = 1 MΩ, C = 1 μF है। Vᵢ(t) एवं Vₒ(t) क्रमशः निवेश बोल्टता और निर्गत बोल्टता है। [चित्र: R₁ निवेश के श्रेणी में; निर्गत C और R₂ की श्रेणी-शाखा के पार लिया गया है।] बहुत उच्च आवृत्तियों पर इस संजाल द्वारा दी जाने वाली शीणता को dB में ज्ञात कीजिए : 10 अंक

(b)

एक प्रतिरोधक विकृति मापी (स्ट्रेन गेज), जिसका गेज गुणक 2·2 है, को एक आयताकार स्टील सलाख (बार), जिसका प्रत्यास्था मॉडुलस, E = 205×10⁶ kN/m² है, पर सीमेंटिकृत किया गया है। स्टील सलाख की चौड़ाई एवं मोटाई क्रमशः 3·5 cm और 0·55 cm है। 12 kN का एक अक्षीय बल लगाया जाता है। यदि विकृति मापी का अभीष्ट (नॉमिनल) प्रतिरोध 100 Ω है, तो विकृति मापी के प्रतिरोध में परिवर्तन निकालिए। 10 अंक

(c)

एक त्रि-कला, 50 Hz, 415 V सप्लाई 250 kW शक्ति, 0·8 पश्चता शक्ति गुणक पर प्रदान करती है। पार्श्वपथ धारिताओं के संस्थापन द्वारा लाइन शक्ति गुणक को 0·9 पश्चता तक सुधार किया जाना वांछित है। यदि उन पार्श्वपथ धारिताओं को डेल्टा में जोड़ा गया हो, तो धारिता ज्ञात कीजिए। 10 अंक

(d)

द्वि-आधारी डाटा को यौगिक सफेद गॉसियन रव (ए० डब्ल्यू० जी० एन०) चैनल पर 5 किलोबिट्स प्रति सेकंड की बिट दर से संचारित किया जाता है। चैनल के लिए एकल-पार्श्व शक्ति स्पेक्ट्रम घनत्व 10⁻⁷ W/Hz है। असंसक्त (नॉन-कोहरेंट) लंबिक द्वि-आधारी एफ० एस० के०, जिसकी अधिक आवृत्ति संकेतन टोन 1 MHz है, को उपयोग में लिया गया है। बिट ऊर्जा, E_b = 2×10⁻⁶ J है। न्यूनतम आवश्यक बैंड-चौड़ाई एवं औसतन बिट त्रुटि प्रायिकता ज्ञात कीजिए। 10 अंक

(e)

एक त्रि-कला, Δ-Y जुड़े हुए, 30 MVA, 33/11 kV, विभेदक रिले रक्षण सहित परिणामित्र लीजिए। यदि प्राइमरी की तरफ CT अनुपात 500 : 5A एवं सेकंडरी की तरफ CT अनुपात 2000 : 5A है, तो ऐसे दोष (फॉल्ट), जो परिणामित्र की नियत धारा की 200% तक धारा ले रहे हैं, उनके लिए रिले धारा अवस्थापन (सेटिंग) ज्ञात कीजिए। 10 अंक

Q5 of the 2023 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2023 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.

(a) A compensator network with an input port on the left and an output port on the right. A continuous reference wire runs along the bottom between the bottom input terminal and bottom output terminal. The input voltage across the input terminals is labeled V_i(t) with a vertical double-headed arrow. From the top input terminal, a horizontal series branch contains a resistor R_1 = 3 MΩ. After R_1, the line connects to a node that continues horizontally to the top output terminal, where the output voltage is labeled V_o(t) with a vertical double-headed arrow. From the node between R_1 and the output terminal, a vertical shunt branch connects to the bottom reference wire, consisting of a capacitor C = 1 μF in series above a resistor R_2 = 1 MΩ.

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) High-frequency attenuation of the compensator network

From the figure, R₁ is in series with the input, and the output is taken across the branch of C in series with R₂. So V_o/V_i = (R₂ + 1/sC)/(R₁ + R₂ + 1/sC) = (1 + sR₂C)/(1 + s(R₁ + R₂)C). This is a lag network with zero at 1/(R₂C) = 1 rad/s and pole at 1/((R₁ + R₂)C) = 0.25 rad/s.

At very high frequency the capacitor is a short circuit, so V_o/V_i → R₂/(R₁ + R₂) = 1/(3 + 1) = 0.25. Attenuation = 20 log₁₀(1/0.25) = 20 log₁₀ 4 = 12.04 dB.

Diagram: Lag network

(b) Strain gauge

Area A = 0.035 × 0.0055 = 1.925 × 10⁻⁴ m². Stress σ = F/A = 12 000/1.925 × 10⁻⁴ = 62.34 × 10⁶ N/m². With E = 205 × 10⁶ kN/m² = 205 × 10⁹ N/m², strain ε = σ/E = 62.34 × 10⁶/205 × 10⁹ = 3.04 × 10⁻⁴.

ΔR/R = G ε, so ΔR = 2.2 × 3.04 × 10⁻⁴ × 100 = 0.0669 Ω.

(c) Power factor correction

P = 250 kW. At cos φ₁ = 0.8, φ₁ = 36.87° and Q₁ = 250 tan 36.87° = 187.5 kVAr. At cos φ₂ = 0.9, φ₂ = 25.84° and Q₂ = 250 tan 25.84° = 121.1 kVAr. The capacitor bank must supply Q_c = 187.5 − 121.1 = 66.4 kVAr.

For a delta bank each capacitor has the line voltage 415 V across it, and carries Q_c/3 = 22.14 kVAr. Since Q = V²ωC, C = 22 140/(415² × 2π × 50) = 4.09 × 10⁻⁴ F = 409 µF per phase.

(d) Non-coherent orthogonal binary FSK

Given: R_b = 5 kb/s, N₀ = 10⁻⁷ W/Hz, E_b = 2 × 10⁻⁶ J, so E_b/N₀ = 20.

For orthogonal non-coherent FSK, the tone spacing is 1/T_b = R_b = 5 kHz. With the higher tone at 1 MHz, the tones are 1 MHz and 995 kHz. The minimum bandwidth is M R_s = 2R_b = 10 kHz, on the Nyquist basis; with null-to-null spectra the width is 5 + 2 × 5 = 15 kHz.

Average bit error probability P_e = ½ exp(−E_b/2N₀) = ½ exp(−10) = 2.27 × 10⁻⁵.

(e) Differential protection of the transformer

The rated primary (33 kV, delta) current is 30 × 10⁶/(√3 × 33 × 10³) = 524.9 A, and the rated secondary (11 kV, star) current is 30 × 10⁶/(√3 × 11 × 10³) = 1574.6 A.

The CTs on the delta side are connected in star, and the CTs on the star side in delta, so the phase shift is compensated. Primary CT secondary current = 524.9/100 = 5.25 A. Secondary CT phase current = 1574.6/400 = 3.94 A, so the line current into the relay is √3 × 3.94 = 6.82 A.

For a fault drawing 200% of rated current, the relay operating current is 2 × 5.25 = 10.5 A, on the ratio 100:1 of the feeding side. The mismatch between 5.25 A and 6.82 A, a ratio of 0.77, must be removed by relay taps or an auxiliary CT so that no spill current flows on through-load.

How this answer will be evaluated

What earns full marks: All parts show complete circuit/model setup, correct derivations, and final answers with units.

Evaluation rubric

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

  1. (a) Find the attenuation in dB of the network at very high frequencies. 10 marks

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

    Must cover

    • Transfer function (1 + sR2C)/(1 + s(R1+R2)C)
    • Capacitor short at high frequency
    • Gain R2/(R1+R2) = 0.25
    • Attenuation 12.04 dB

    Loses marks

    • Reporting the gain as positive dB.

    Earns more

    • Marks for the transfer function, the limit and the dB value.

    Extra mark

    • Naming it a lag network with pole and zero values.
  2. (b) Find the change in resistance of the strain gauge. 10 marks

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

    Must cover

    • Stress = F/A = 62.34 MPa
    • Strain = stress/E = 3.04e-4
    • dR = G x strain x R
    • dR = 0.0669 ohm

    Loses marks

    • Unit slip between kN/m2 and N/m2.

    Earns more

    • Marks for stress, strain and the resistance change with units.

    Extra mark

    • Converting E from kN/m2 correctly.
  3. (c) Calculate the delta-connected capacitance for power factor correction. 10 marks

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

    Must cover

    • Q1 = 187.5 kVAr at 0.8
    • Q2 = 121.1 kVAr at 0.9
    • Qc = 66.4 kVAr, 22.14 kVAr per phase
    • C = 409 uF with line voltage across each capacitor

    Loses marks

    • Using phase voltage 240 V across delta capacitors.

    Earns more

    • Marks for the reactive powers, the bank rating and C.

    Extra mark

    • Noting C for star would be three times larger.
  4. (d) Find the minimum bandwidth and average bit error probability of non-coherent FSK. 10 marks

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

    Must cover

    • Eb/N0 = 20
    • Tone spacing 1/Tb = 5 kHz
    • Minimum bandwidth 2Rb = 10 kHz
    • Pe = 0.5 exp(-Eb/2N0) = 2.27e-5

    Loses marks

    • Using the coherent FSK formula.

    Earns more

    • Marks for Eb/N0, bandwidth and error probability.

    Extra mark

    • Noting the 15 kHz null-to-null width.
  5. (e) Compute the relay current setting for faults up to 200% of rated current. 10 marks

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

    Must cover

    • Rated currents 524.9 A and 1574.6 A
    • CTs on delta side in star, on star side in delta
    • Secondary currents 5.25 A and 6.82 A
    • Relay current 10.5 A at 200%

    Loses marks

    • Connecting the CTs on both sides in the same way as the windings.

    Earns more

    • Marks for rated currents, CT connections and the relay current.

    Extra mark

    • Noting the tap or auxiliary CT needed to remove the mismatch.
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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