Electrical Engineering 2022 Paper II 50 marks Solve

Paper II — Q3

(a) (i) The figure below shows a schematic diagram of shunt ohmmeter. It consists of a battery in series with an adjustable…

(a)
(i)

The figure below shows a schematic diagram of shunt ohmmeter. It consists of a battery in series with an adjustable resistance R1 and a d.c. measuring PMMC ammeter. The unknown resistance Rx to be measured is connected across the terminals A-B. S is the off-on switch. Write down its design equations. What are your observations when the terminals A-B are shorted and open-circuited? [Figure: battery E, adjustable R1 and switch S in series, the basic PMMC meter (Rm, current Im) connected between terminal A and terminal B, and the unknown Rx also connected across A-B, so the meter and Rx are in parallel and share the battery current I1] 10 marks

(ii)

Determine the value of the resistor in series with the battery to adjust the full-scale deflection and the point (in %) of full scale at which 100 Ω will be marked on the scale. Consider the e.m.f. of internal battery as 2 volts, full-scale current of 2 mA and an internal resistance of 20 Ω. 10 marks

(b)

Consider the system

x̄ = [0 16 0 0; -1 -8 0 0; 4 8 1 15; 0 0 3 -3] x̄ + [0; 0; 1; 0] u

(i)

Find the eigenvalues of this system.

(ii)

Find the controllable and uncontrollable modes of this system.

(iii)

Show that there are an infinite number of feedback gains K that will relocate the modes of the system to -5, -3, -4, -4.

(iv)

Find the unique matrix K̄ that achieves these pole locations and prevents initial conditions on the uncontrollable part of the system from ever affecting the controllable part. 20 marks

(c)
(i)

Explain the use of MAR in interfacing the external memory to processor.

(ii)

Name various 8-bit and 16-bit internal registers in microprocessor 8085 and also, indicate which 8-bit registers can be paired to act as 16-bit register in several instructions. 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

नीचे चित्र में एक शंट ऑहमीटर का योजनाबद्ध आरेख प्रदर्शित है। इसमें एक बैटरी के श्रेणीक्रम में एक समायोज्य प्रतिरोध R1 तथा एक डी. सी. मापक पी. एम. सी. ए. एमीटर जुड़े हुए हैं। अज्ञात प्रतिरोध Rx, जिसका मापन करना है, छोरों A-B के आर-पार लगा है। S बंद-चालू (ऑफ-ऑन) कुंजी है। इसका अभिकल्प समीकरण लिखिए। छोर A-B के जुड़े हुए एवं खुले परिपथ की स्थिति में आपका क्या प्रेक्षण है? [figure] (10 अंक)

(ii)

पूर्ण-पैमाना विक्षेप के समायोजन के लिए बैटरी के श्रेणीक्रम में प्रतिरोध का मान तथा पूर्ण पैमाना का वह बिंदु (प्रतिशत में), जहाँ पैमाने पर 100 Ω अंकित किया जाएगा, ज्ञात कीजिए। आन्तरिक बैटरी का वि. वा. ब. 2 वोल्ट, पूर्ण-पैमाना धारा 2 mA तथा आन्तरिक प्रतिरोध 20 Ω मान लीजिए। (10 अंक)

(b)

नीचे प्रदशित प्रणाली पर विचार कीजिए:

x̄ = [0 16 0 0; -1 -8 0 0; 4 8 1 15; 0 0 3 -3] x̄ + [0; 0; 1; 0] u

(i)

इस प्रणाली के आइगेन मानों को ज्ञात कीजिए।

(ii)

इस प्रणाली की नियंत्रणीय तथा अनियंत्रणीय विधाएँ ज्ञात कीजिए।

(iii)

प्रदर्शित कीजिए कि पुनर्निवेश लब्धि K के ऐसे अंशक्य मान होंगे, जो प्रणाली की विधाओं को -5, -3, -4, -4 पर पुनर्स्थापित करेंगे।

(iv)

अन्य अत्यवश्यक K̄ ज्ञात कीजिए, जो इन ध्रुवों के स्थान को प्राप्त करके प्रणाली के अनियंत्रणीय अंश की आरम्भिक शर्तों को नियंत्रणीय अंश पर प्रभाव डालने से सदा रोक देता है। (20 अंक)

(c)
(i)

संसाधक को बाह्य मेमोरी से अंतरपुष्ट (इंटरफेस) करने में एम. ए. आर. का उपयोग समझाइए।

(ii)

सूक्ष्म-संसाधक 8085 में विभिन्न आन्तरिक 8-बिट तथा 16-बिट पंजियों को नामित कीजिए और यह भी इंगित कीजिए कि कौन-सी 8-बिट पंजियाँ युग्मित होकर विभिन्न अनुदेशों में 16-बिट पंजी की तरह कार्य कर सकती हैं। (10 अंक)

Q3 of the 2022 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2022 Electrical Engineering paper

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)(i) Shunt ohmmeter: E in series with adjustable R1 and switch S, basic meter Rm across terminals A-B, unknown Rx across A-B in parallel with the meter.

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)(i) Shunt ohmmeter: design equations

In the shunt ohmmeter the battery E, the adjustable resistance R₁ and the switch S are in series, and the basic PMMC meter (resistance Rm, full-scale current Im) is connected directly across the output terminals A–B, in parallel with the unknown Rx. The meter therefore shares the battery current I₁ with Rx.

With A–B open, all of I₁ flows through the meter. R₁ is adjusted for full-scale deflection:

Ifs = E/(R₁ + Rm), so R₁ = E/Ifs − Rm.

With Rx connected, the battery sees R₁ in series with Rm ∥ Rx, and the meter current is the current-divider share of I₁:

I₁ = E/[R₁ + Rm Rx/(Rm + Rx)], Im = I₁ Rx/(Rm + Rx) = E Rx/(R₁Rm + R₁Rx + RmRx).

The deflection as a fraction of full scale is Im/Ifs = Rx(R₁ + Rm)/(R₁Rm + R₁Rx + RmRx), and half-scale deflection occurs at Rh = R₁Rm/(R₁ + Rm), the parallel combination of the meter and the series resistance. These are the design equations; the scale is marked from them.

Observations: with A–B shorted, Rx = 0, the meter is short-circuited and reads zero, so the zero-ohm mark is at the left of the scale (no deflection). With A–B open, Rx = ∞, the meter carries the full battery current and reads full scale, so infinity is at the right of the scale. The scale therefore runs the opposite way to a series ohmmeter, and it is non-linear, crowded towards the high-resistance end. The switch S is essential because the battery would otherwise discharge continuously through R₁ and the meter even with nothing connected; it also allows the meter to be set to zero (open circuit gives full scale, so the zero adjustment is made with A–B open).

(a)(ii) Series resistance and the 100 Ω mark

E = 2 V, Ifs = 2 mA, Rm = 20 Ω.

R₁ = E/Ifs − Rm = 2/0.002 − 20 = 1000 − 20 = 980 Ω.

For Rx = 100 Ω:

Im = E Rx/(R₁Rm + R₁Rx + RmRx) = 2 × 100/(980 × 20 + 980 × 100 + 20 × 100) = 200/(19,600 + 98,000 + 2,000) = 200/119,600 = 1.672 mA.

Im/Ifs = 1.672/2 = 0.836, so 100 Ω is marked at 83.6% of full scale. As a check, the half-scale resistance is Rh = 980 × 20/1000 = 19.6 Ω, and 100 Ω, being much larger than Rh, lies well up the scale, as found.

(b) The state-space system

A = [0 16 0 0; −1 −8 0 0; 4 8 1 15; 0 0 3 −3], B = [0; 0; 1; 0].

A is block lower-triangular: the first two states (x₁, x₂) evolve on their own through A₁₁ = [0 16; −1 −8], while (x₃, x₄) are driven by A₂₂ = [1 15; 3 −3], by the coupling A₂₁ = [4 8; 0 0] from the first pair, and by the input through B₂ = [1; 0].

(i) Eigenvalues. det(λI − A) = det(λI − A₁₁) det(λI − A₂₂). For A₁₁: λ² + 8λ + 16 = (λ + 4)², roots −4, −4. For A₂₂: λ² + 2λ − 48 = 0, roots λ = (−2 ± 14)/2 = 6 and −8. The eigenvalues are −4, −4, 6, −8; the system is unstable because of the mode at +6.

(ii) Controllable and uncontrollable modes. The input enters only x₃, and nothing in (x₃, x₄) feeds back into (x₁, x₂), because A₁₂ = 0. The pair (x₁, x₂) is therefore unaffected by u: the modes −4, −4 are uncontrollable. For the (x₃, x₄) subsystem the controllability matrix is [B₂ A₂₂B₂] = [1 1; 0 3], of rank 2, so the modes 6 and −8 are controllable.

(iii) Infinitely many gains. With state feedback u = −Kx, K = [k₁ k₂ k₃ k₄], only the third row of A changes:

A − BK = [0 16 0 0; −1 −8 0 0; 4−k₁ 8−k₂ 1−k₃ 15−k₄; 0 0 3 −3].

The closed loop keeps the block lower-triangular form, so its eigenvalues are those of A₁₁, still −4, −4, together with those of A₂₂ − B₂K₂, K₂ = [k₃ k₄]:

det(λI − (A₂₂ − B₂K₂)) = λ² + (2 + k₃)λ + (3k₃ + 3k₄ − 48).

The uncontrollable modes stay at −4, −4 whatever K is; the required set −5, −3, −4, −4 is reachable precisely because its uncontrollable part is already at −4, −4. Matching λ² + (2 + k₃)λ + (3k₃ + 3k₄ − 48) to (λ + 5)(λ + 3) = λ² + 8λ + 15 gives k₃ = 6 and 18 + 3k₄ − 48 = 15, k₄ = 15. The gains k₁ and k₂ do not appear in the characteristic polynomial, so every K = [k₁ k₂ 6 15] places the poles at −5, −3, −4, −4: there are infinitely many.

(iv) The unique K̄. The coupling from the uncontrollable states into the controllable ones is the block A₂₁ − B₂K₁, whose first row is [4 − k₁, 8 − k₂] and second row [0 0]. Choosing k₁ = 4, k₂ = 8 makes it zero, so that initial conditions on (x₁, x₂) never reach (x₃, x₄). Hence

K̄ = [4 8 6 15],

and the closed-loop matrix becomes [0 16 0 0; −1 −8 0 0; 0 0 −5 0; 0 0 3 −3] in its third row, fully decoupled, with eigenvalues −4, −4, −5, −3 as required.

(c)(i) The memory address register

The MAR holds the 16-bit address of the memory location the processor is about to read or write. During a machine cycle the address is transferred from the program counter (for an opcode fetch), the stack pointer (for PUSH, POP, CALL, RET) or the HL pair (for MOV M and other memory-referencing instructions) into the MAR, whose outputs drive the address bus. In the 8085 the higher byte appears on A15–A8 for the whole machine cycle, while the lower byte appears on the multiplexed lines AD7–AD0 only in T1; the ALE pulse latches it into an external 8-bit latch (74LS373), which thus acts as the external half of the MAR and holds the low address steady while AD7–AD0 carry data in T2–T3. The address decoder then selects the chip, and RD or WR strobes the transfer. The MAR is what keeps the address stable, and the memory selected, for the whole of the read or write.

(c)(ii) Registers of the 8085

8-bit registers: the accumulator A; the six general-purpose registers B, C, D, E, H and L; the flag register F (S, Z, AC, P, CY); and, not accessible to the programmer, the instruction register and the temporary registers W and Z. 16-bit registers: the program counter PC and the stack pointer SP.

The general-purpose registers pair up as BC, DE and HL for 16-bit operations: LXI loads a pair, INX and DCX increment and decrement it, PUSH and POP transfer it to and from the stack, DAD adds BC, DE, HL or SP to HL, XCHG exchanges DE and HL, and LDAX and STAX use BC or DE as a memory pointer. HL is the memory pointer M for the whole MOV/ADD/ANA family, and A with F forms the pair PSW for PUSH PSW and POP PSW.

How this answer will be evaluated

What earns full marks: Complete derivations, correct calculations, clear controllability analysis, accurate register listing.

Evaluation rubric

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

  1. (a)(i) Design equations for shunt ohmmeter and observations for short/open terminals. 10 marks

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

    Must cover

    • Redraw circuit with polarities and reference directions
    • Derive equation for short-circuit current (Isc)
    • Derive equation for current with Rx connected
    • State observations for shorted and open-circuited terminals

    Loses marks

    • Formula application without equivalent circuit
    • Sign errors in KCL/KVL equations

    Earns more

    • Explicitly define full-scale deflection condition
    • Show relationship between Rx and meter deflection

    Extra mark

    • Include a labelled equivalent circuit diagram
  2. (a)(ii) Value of series resistor R1 and scale point for 100 Ω. 10 marks

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

    Must cover

    • Given: E=2V, Ifs=2mA, Rm=20Ω
    • Calculate R1 for full-scale deflection
    • Calculate current for Rx=100Ω
    • R1 = 980 Ω, Im at 100 Ω = 1.672 mA, i.e. 83.6% of full scale

    Loses marks

    • Arithmetic errors in resistance calculation
    • Missing units in final answer

    Earns more

    • Show substitution steps clearly
    • State units for all intermediate values

    Extra mark

    • Comment on linearity of the scale
  3. (b) Eigenvalues, controllability, feedback gains, and unique K matrix. 20 marks

    analyse— intro → causes → effects → stakeholders/linkages → way forward

    Must cover

    • Eigenvalues −4, −4, 6, −8
    • Modes −4, −4 uncontrollable, 6 and −8 controllable
    • Any K = [k1 k2 6 15] gives poles −5, −3, −4, −4
    • Unique K̄ = [4 8 6 15] removes the coupling from the uncontrollable states

    Loses marks

    • Incorrect eigenvalue calculation
    • Confusing controllable and uncontrollable modes

    Earns more

    • Use PBH test or rank condition for controllability
    • Show characteristic equation derivation

    Extra mark

    • Provide a block diagram of the system
  4. (c) Use of MAR and 8085 registers with pairing capabilities. 10 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Explain MAR function in memory interfacing
    • List 8-bit registers in 8085
    • List 16-bit registers in 8085
    • Indicate which 8-bit registers pair to 16-bit

    Loses marks

    • Omitting any standard 8-bit register
    • Incorrect pairing of registers

    Earns more

    • Mention specific instructions using paired registers
    • Clarify role of SP and PC as 16-bit

    Extra mark

    • Include a register map diagram
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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