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Electrical Engineering 2021 Paper II 50 marks Compulsory Explain

Paper II — Q5

(a) Explain the ratio error and phase angle error of current transformer. 10 marks (b) The two top rows of a Routh table of a…

(a)

Explain the ratio error and phase angle error of current transformer. 10 marks

(b)

The two top rows of a Routh table of a characteristic polynomial is given in the table. Determine the roots of the characteristic equation which lie in the left half s-plane. Complete the remaining rows of the table. 10 marks

(c)

A pulse is applied to a piezo-electric transducer for a time T. Prove that in order to keep the undershoot of the response to a value within 5%, the value of time constant should be approximately 20T. 10 marks

(d)
(i)

A discrete memoryless source (DMS) has five symbols x₁, x₂, x₃, x₄ and x₅ with P(x₁) = 0·4, P(x₂) = 0·19, P(x₃) = 0·16, P(x₄) = 0·15 and P(x₅) = 0·1. Construct a Shannon Fano code for the source and calculate the efficiency of the code.

(ii)

Repeat for Huffman code. Compare the results of (i) and (ii). 10 marks

(e)

List the functional classification of 8085 instruction set. Give one example for each class. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

विद्युत धारा परिणामित्र की अनुपातिक त्रुटि व कला कोण त्रुटि की व्याख्या करें । 10

(b)

एक अभिलक्षण बहुपद (कैरेक्टरिस्टिक पॉलिनोमियल) की रूथ सारणी की सबसे ऊपर की दो पंक्तियाँ नीचे दर्शायी गई हैं । s-तल के अर्ध बाम में स्थित अभिलक्षण समीकरण के मूलों का निर्धारण करें व सारणी की शेष पंक्तियों को पूर्ण करें । 10

(c)

एक दाब विद्युत पारातंत्र पर T समय के लिए एक स्पंदन अनुप्रयुक्त किया गया है । सिद्ध करें कि समय स्थिरांक का अनुमानित मान 20T होगा यदि प्रतिक्रिया का अधोचरम (अंडरशूट) मान 5% तक सीमित हो । 10

(d)
(i)

एक असतत स्मृतिहीन स्रोत (DMS) पांच प्रतीक चिह्न x₁, x₂, x₃, x₄, x₅ हैं जहाँ कि; P(x₁) = 0·4, P(x₂) = 0·19, P(x₃) = 0·16, P(x₄) = 0·15, P(x₅) = 0·1 है । इस स्रोत के लिए शैनन फैनो कूट (कोड) का निर्धारण करें व कूट (कोड) की दक्षता की गणना करें ।

(ii)

इस स्रोत के लिए हफमैन कूट का भी निर्धारण करें व (i) व (ii) के परिणामों की तुलना करें । 10

(e)

8085 सूक्ष्म संसाधक के अनुदेश समुच्चय को कार्यात्मक वर्गीकरण के अनुसार सूचीबद्ध करें । प्रत्येक वर्ग का एक उदाहरण लिखें । 10

Q5 of the 2021 UPSC Mains Electrical Engineering Paper II, 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) A table with two rows and three columns. The first column contains the row labels 's^4' and 's^3'. The second row of the table contains the values '1', '10', and '24'. The third row of the table contains the values '5', '20', and an empty cell.

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) Ratio error and phase angle error of a CT A CT needs an excitation current I₀ to magnetise its core and supply core loss, so I_p = nI_s + I₀ as phasors (n = turns ratio, K_n = nominal ratio). Ratio error = (K_nI_s − I_p)/I_p × 100%: I_p is slightly larger than K_nI_s, so the error is usually negative. It arises mainly from the core-loss component I_c of I₀, which is in phase with nI_s for a resistive burden. Phase angle error δ is the angle by which the reversed secondary current −I_s leads I_p (ideally 180° between them). It arises mainly from the magnetising component I_μ, which is in quadrature: δ ≈ I_μ/(nI_s) rad. Both grow with burden and with core saturation, since both raise I₀, and they matter for metering and relay accuracy. They are reduced by a high-permeability core, a low burden and turns compensation.

(b) Routh table The polynomial is s⁴ + 5s³ + 10s² + 20s + 24. s² row: (5 × 10 − 20)/5 = 6 and (5 × 24)/5 = 24. s¹ row: (6 × 20 − 5 × 24)/6 = 0 and 0, a row of zeros, so form the auxiliary polynomial A(s) = 6s² + 24 and use dA/ds = 12s: s⁴: 1, 10, 24; s³: 5, 20; s²: 6, 24; s¹: 12; s⁰: 24. There is no sign change, so no right-half-plane roots. A(s) = 0 gives s = ±j2 on the imaginary axis, and dividing, (s² + 4)(s² + 5s + 6) = 0. The left-half-plane roots are s = −2 and s = −3; the system is marginally stable.

(c) Undershoot of the piezoelectric transducer The transducer behaves as a charge source q = dF across C and R in parallel, so its response to a step force F₀ is v = (dF₀/C)e^(−t/τ) with τ = RC (a high-pass response). A pulse of length T is a step at t = 0 minus a step at t = T: v(t) = (dF₀/C)[e^(−t/τ) − e^(−(t − T)/τ)] for t > T. At t = T⁺ the output is (dF₀/C)(e^(−T/τ) − 1), an undershoot of relative size 1 − e^(−T/τ) ≈ T/τ for T ≪ τ. For this to be within 5%: T/τ ≤ 0.05, i.e. τ ≥ 20T. (Exactly, −ln 0.95 = 0.0513, so τ ≥ 19.5T.)

(d) Source coding Entropy H = Σp log₂(1/p) = 0.5288 + 0.4552 + 0.4230 + 0.4105 + 0.3322 = 2.150 bits/symbol. (i) Shannon–Fano. The split {x₁, x₂} | {x₃, x₄, x₅} (0.59 : 0.41) is closest to equal. Then x₁ = 00, x₂ = 01; {x₃} | {x₄, x₅} gives x₃ = 10, x₄ = 110, x₅ = 111. L = 0.4 × 2 + 0.19 × 2 + 0.16 × 2 + 0.15 × 3 + 0.1 × 3 = 2.25 bits, so η = 2.150/2.25 = 95.5%. (ii) Huffman. Merge 0.1 + 0.15 = 0.25; 0.16 + 0.19 = 0.35; 0.25 + 0.35 = 0.60; 0.40 + 0.60 = 1. Code: x₁ = 0, and x₂, x₃, x₄, x₅ get 3-bit codes (for example 100, 101, 110, 111 by tree position). L = 0.4 + 0.6 × 3 = 2.20 bits, so η = 2.150/2.20 = 97.7%. Huffman is shorter by 0.05 bit/symbol and more efficient, since it is optimal for the given probabilities.

(e) 8085 instruction classes Data transfer: MOV A, B (also MVI, LDA, STA, PUSH, POP, IN, OUT); arithmetic: ADD B; logical: ANA B (also CMP, rotate); branching: JMP 2000H (also CALL, RET); machine control: HLT (also NOP, EI, DI). Stack, I/O and machine-control instructions are sometimes counted as separate classes.

How this answer will be evaluated

What earns full marks: All parts answered with correct method, complete working, and clear interpretation.

Evaluation rubric

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

  1. (a) Define and explain the ratio error and phase angle error of a current transformer. 10 marks

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

    Must cover

    • Define ratio error (K_nI_s − I_p)/I_p × 100% and phase angle error (angle between I_p and the reversed I_s).
    • Explain the causes through the excitation current I₀: core-loss component and magnetising component.
    • State the effect on metering and relay accuracy and how burden and saturation increase them.

    Loses marks

    • Confusing ratio error with phase error.
    • Omitting the 180° reference for the phase angle.

    Earns more

    • Effect of burden impedance mentioned.
    • Effect of core saturation mentioned.

    Extra mark

    • Approximate formulas such as δ ≈ I_μ/(nI_s).
  2. (b) Complete the Routh table of s⁴ + 5s³ + 10s² + 20s + 24 and find the left-half-plane roots. 10 marks

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

    Must cover

    • Compute the s² row (6, 24) and recognise the zero s¹ row, forming A(s) = 6s² + 24.
    • Complete the table using dA/ds = 12s (s¹: 12; s⁰: 24) and note no sign change.
    • Find s = ±j2 from A(s) and the left-half-plane roots s = −2 and s = −3.

    Loses marks

    • Arithmetic errors in the table.
    • Wrong interpretation of the zero row or the sign changes.

    Earns more

    • Calculation steps for each entry shown.
    • Stability conclusion (marginally stable) stated.

    Extra mark

    • Factorisation (s² + 4)(s² + 5s + 6) as a check.
  3. (c) Prove that the time constant must be about 20T for the undershoot of a piezoelectric transducer to stay within 5%. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Model the transducer as a charge source with C and R in parallel, giving a step response e^(−t/τ).
    • Write the pulse as a step minus a delayed step and find the undershoot 1 − e^(−T/τ) ≈ T/τ.
    • Set it to 5% to get τ ≈ 20T.

    Loses marks

    • Skipping derivation steps.
    • Algebraic errors in the undershoot expression.

    Earns more

    • Exponential decay term shown.
    • Approximation T ≪ τ stated.

    Extra mark

    • Sketch of the pulse response.
  4. (d) Shannon–Fano and Huffman codes for the five-symbol source, their efficiencies and a comparison. 10 marks

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

    Must cover

    • Compute the entropy 2.150 bits and build the Shannon–Fano code with lengths 2, 2, 2, 3, 3 (L = 2.25, η = 95.5%).
    • Build the Huffman code with lengths 1, 3, 3, 3, 3 (L = 2.20, η = 97.7%).
    • Compare: Huffman is shorter and more efficient.

    Loses marks

    • Incorrect code construction.
    • Arithmetic errors in efficiency.

    Earns more

    • Probability sorting and splits shown for Shannon–Fano.
    • Huffman merging steps shown.

    Extra mark

    • Redundancy of each code.
  5. (e) Functional classification of the 8085 instruction set with one example of each class. 10 marks

    enumerate— list the items in order → one line each → no commentary

    Must cover

    • Name the classes: data transfer, arithmetic, logical, branching and machine control (stack and I/O noted).
    • Give a valid example of each (MOV A, B; ADD B; ANA B; JMP 2000H; HLT).
    • Cover all major classes.

    Loses marks

    • Missing a major class.
    • Invalid 8085 instruction examples.

    Earns more

    • Brief description of each class.
    • Extra instructions per class.

    Extra mark

    • Opcodes of the examples.
What "Explain" is asking you to do

Make the working of something clear — what sets it off, what follows from what, and what it produces. Explain is the Commission's mechanism word: it dominates the technical papers and the “explain why” stems, where the marks sit in the causal chain and not in the label.

Structure that answers it

State what it is → the initiating condition → the chain of cause, step by step → an instance where it plays out → what the chain produces

Where marks are lost

Describing what something looks like instead of why it works that way. Naming the stages without linking them reads as description too.

All UPSC directive words, compared →

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