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Electrical Engineering 2023 Paper II 50 marks Compulsory Solve

Paper II — Q1

(a) The figure shows a unity feedback system with open-loop transfer function G(s) = K/((s + 1)(s + 2)). The steady-state value…

(a)

The figure shows a unity feedback system with open-loop transfer function G(s) = K/((s + 1)(s + 2)). The steady-state value of the unit step response c(t) is 0·8. Determine the maximum overshoot in the response c(t) : 10 marks

(b)

A circuit breaker is rated as 2500 A, 1500 MVA, 33 kV, 3 sec, 3-phase, oil circuit breaker. Determine its rated normal current, breaking current, making current and short-time rating (current). 10 marks

(c)

An audio signal, whose bandwidth is 15 kHz, is to be digitized using PCM. Uniform quantization with 1024 levels and binary encoding are assumed. Determine the minimum sampling rate. If the actual sampling rate is 20% excess of the minimum rate, determine the minimum permissible bit rate. 10 marks

(d)

Briefly explain the following logical instructions of 8085 microprocessor : (i) ANA M (ii) XRA M (iii) CMC (iv) STC (v) RRC 10 marks

(e)

In a three-phase 400 km long transmission line, the conductors are spaced at the corners of an equilateral triangle of side 5 m. The diameter of each conductor is 3 cm. Calculate the capacitance per phase of the 400 km long conductor. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

चित्र में एक इकाई पुनर्निवेश तंत्र दिखाया गया है, जिसका खुला-पाश अंतरण फलन G(s) = K/((s + 1)(s + 2)) है। इकाई पद अनुक्रिया (रेस्पोंस) c(t) का स्थायी-दशा मान 0·8 है। अनुक्रिया c(t) में अधिकतम ओवरशूट ज्ञात कीजिए : (10 अंक)

(b)

एक परिपथ विचोजक को 2500 A, 1500 MVA, 33 kV, 3 sec, 3-कला, ऑयल परिपथ विचोजक की तरह निर्धारित किया गया है। उसकी निर्धारित सामान्य धारा, विचोजन धारा, संयोजन धारा एवं लघु-समय रेटिंग (धारा) ज्ञात कीजिए। (10 अंक)

(c)

एक श्रव्य संकेत, जिसकी बैंड-चौड़ाई 15 kHz है, को पी० सी० एम० का उपयोग करते हुए अंकिक बनाना है। 1024 स्तरों (levels) के साथ एकसमान कांटन एवं द्वि-आधारी कोडन मान लीजिए। न्यूनतम प्रतिचयन दर (मिनिमम सैम्पलिंग रेट) निर्धारित कीजिए। यदि वास्तविक प्रतिचयन दर, न्यूनतम दर से 20% ज्यादा है, तो न्यूनतम अनुमत्य बिट दर ज्ञात कीजिए। (10 अंक)

(d)

8085 सूक्ष्म-संसाधित्र (माइक्रोप्रोसेसर) के निम्नलिखित तार्किक निर्देशों की संक्षिप्त व्याख्या कीजिए : (i) ANA M (ii) XRA M (iii) CMC (iv) STC (v) RRC (10 अंक)

(e)

एक त्रि-कला (श्री-फेज़) 400 km लम्बी संचरण लाइन में, चालकों (कण्डक्टरों) को 5 m भुजा वाले एक समबाहु त्रिभुज के कोनों के अन्तराल पर रखा गया है। प्रत्येक चालक का व्यास 3 cm है। इस 400 km लम्बे चालक की प्रति कला धारिता की गणना कीजिए। (10 अंक)

Q1 of the 2023 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2023 Electrical Engineering paper
Evaluate your answer to this question → See all 2023 Electrical Engineering questions

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 block diagram of a unity negative feedback control system: An input signal u(t) goes into a summing junction with a positive (+) sign. The error output from the summing junction enters a forward-path transfer function block with transfer function K / ((s + 1)(s + 2)). The output of this block is c(t). A unity feedback path is taken from c(t) and fed back into the summing junction with a negative (-) sign.

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) Maximum overshoot of the unity feedback system

With G(s) = K/((s + 1)(s + 2)) in a unity feedback loop, C(s)/R(s) = K/(s² + 3s + 2 + K). For a unit step, the steady-state value is K/(2 + K) = 0.8, so K = 8.

The characteristic equation is s² + 3s + 10 = 0, so ω_n = √10 = 3.162 rad/s and 2ζω_n = 3, giving ζ = 3/(2 × 3.162) = 0.474 (underdamped).

Maximum overshoot = e^(−πζ/√(1 − ζ²)) = e^(−π × 0.474/0.880) = e^(−1.693) = 0.184, that is 18.4 per cent of the final value. The peak value of c(t) is 0.8 × 1.184 = 0.947.

Diagram: Unity feedback system with G of s equal to 8 over (s+1)(s+2) and its unit step response settling at 0.8 with 18.4 percent overshoot

(b) Circuit breaker ratings

(i) Rated normal current: 2500 A, as stated.

(ii) Breaking current. Breaking capacity = √3 V I, so the rated (symmetrical) breaking current is I_b = 1500 × 10⁶/(√3 × 33 × 10³) = 26.24 kA.

(iii) Making current. It is the peak of the first current loop, taken as 2.55 times the symmetrical breaking current: 2.55 × 26.24 = 66.9 kA (peak).

(iv) Short-time rating. The breaker must carry its rated breaking current for the rated time of 3 s, so the short-time current is 26.24 kA for 3 s.

(c) PCM sampling rate and bit rate

The minimum (Nyquist) sampling rate is f_s = 2 f_m = 2 × 15 kHz = 30 kHz.

With 1024 = 2¹⁰ levels and binary coding, each sample needs n = 10 bits.

If the actual sampling rate is 20 per cent above the minimum, f_s = 1.2 × 30 = 36 kHz, and the minimum permissible bit rate is n f_s = 10 × 36 000 = 360 kbit/s.

(d) 8085 logical instructions

(i) ANA M: the accumulator is ANDed with the byte in the memory location addressed by the HL pair, and the result is stored in the accumulator. The flags S, Z and P are set according to the result, CY is reset and AC is set.

(ii) XRA M: the accumulator is EX-ORed with the memory byte addressed by HL and the result goes to the accumulator. S, Z and P follow the result, and CY and AC are reset. (With the same operands it clears the accumulator.)

(iii) CMC: complements the carry flag, CY ← CY', and no other flag or register is affected.

(iv) STC: sets the carry flag, CY = 1, and affects no other flag.

(v) RRC: rotates the accumulator right by one bit: D₀ goes to both D₇ and the carry flag, and every other bit moves one place to the right. Only CY is affected.

(e) Capacitance of the 400 km line

For a three-phase line with equilateral spacing d = 5 m and conductor radius r = 3/2 = 1.5 cm, the capacitance from line to neutral is C_n = 2πε₀/ln(d/r) per metre.

C_n = 2π × 8.854 × 10⁻¹²/ln(5/0.015) = 5.563 × 10⁻¹¹/ln(333.3) = 5.563 × 10⁻¹¹/5.809 = 9.58 × 10⁻¹² F/m = 9.58 nF/km.

For 400 km: C = 9.58 × 10⁻¹² × 400 × 10³ = 3.83 µF per phase (to neutral).

The effect of the earth and of the transposition of conductors is neglected.

Diagram: Three conductors at the corners of an equilateral triangle of side 5 m, each of diameter 3 cm

How this answer will be evaluated

Framework: Electrical Engineering, Paper 2

What earns full marks: All parts with complete working, correct formulas, units, and interpretation

Evaluation rubric

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

  1. (a) Determine the maximum overshoot of the unity feedback system given a steady-state value of 0.8. 10 marks

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

    Must cover

    • Closed loop K/(s^2 + 3s + 2 + K) and steady state K/(2 + K) = 0.8, K = 8
    • wn = 3.162 rad/s, zeta = 0.474
    • Overshoot exp(-pi zeta / sqrt(1 - zeta^2)) = 18.4 per cent
    • Peak 0.947

    Loses marks

    • Computing overshoot with respect to 1 instead of the final value.

    Earns more

    • Marks for K, damping ratio and the overshoot.

    Extra mark

    • Peak value of c(t).
  2. (b) Determine the rated normal current, breaking current, making current and short-time rating of the breaker. 10 marks

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

    Must cover

    • Normal current 2500 A
    • Breaking current 26.24 kA from MVA/(sqrt3 kV)
    • Making current 2.55 times breaking = 66.9 kA
    • Short-time rating 26.24 kA for 3 s

    Loses marks

    • Using the MVA formula without sqrt3.

    Earns more

    • Marks for each of the four ratings.

    Extra mark

    • Stating the making current is a peak value.
  3. (c) Determine the minimum sampling rate and the minimum permissible bit rate of the PCM system. 10 marks

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

    Must cover

    • Nyquist rate 30 kHz
    • n = 10 bits for 1024 levels
    • Actual rate 36 kHz
    • Bit rate 360 kbit/s

    Loses marks

    • Using 15 kHz as the sampling rate.

    Earns more

    • Marks for each step and the final bit rate.

    Extra mark

    • Stating the bandwidth requirement.
  4. (d) Explain the 8085 instructions ANA M, XRA M, CMC, STC and RRC. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • ANA M: AND with memory at HL, result in A, flags set
    • XRA M: XOR with memory at HL, CY and AC reset
    • CMC complements carry; STC sets carry
    • RRC rotates A right, D0 to D7 and CY

    Loses marks

    • Confusing RRC with rotate through carry.

    Earns more

    • Marks for each instruction's operation and flag effect.

    Extra mark

    • Noting the HL addressing and accumulator result.
  5. (e) Calculate the capacitance per phase of the 400 km line. 10 marks

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

    Must cover

    • Line-to-neutral capacitance 2 pi eps0 / ln(d/r)
    • r = 1.5 cm and d = 5 m
    • 9.58 nF/km
    • 3.83 microfarad for 400 km

    Loses marks

    • Using the diameter instead of radius.

    Earns more

    • Marks for the formula, substitution and the total.

    Extra mark

    • Stating neglect of earth effect.
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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