Chat with us
Electrical Engineering 2021 Paper I 50 marks Compulsory Solve

Paper I — Q1

(a) In Figure 1(a) shown below, the two-port network is characterized in terms of y-parameters with y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 ×…

(a)

In Figure 1(a) shown below, the two-port network is characterized in terms of y-parameters with y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 × 10⁻³ S and y₁₂ = y₂₁ = 0. Find the voltage across 200 Ω load. 10 marks

(b)

For the signal shown in Figure 1(b), calculate the total energy of the signal X(t). Also sketch y(t) = X(10t – 5). 10 marks

(c)
(i)

A 220 V dc shunt motor has armature resistance Rₐ = 0·13 Ω, field resistance Rf = 250 Ω and rotational loss 230 W. On full-load, the line current is 9·5 A with the motor running at 1440 rpm. Determine the following: The mechanical power developed

(ii)

The power output

(iii)

The load torque

(iv)

The full-load efficiency 10 marks

(d)

For the transistor circuit shown in Figure 1(d), determine the value of reverse saturation current, I_S, that would give a collector current of 1 mA, if β = 80, V_A = ∞ and V_T = 26 mV at T = 300 K. 10 marks

(e)

Consider the four variables logic function defined as follows: F (A, B, C, D) = ĀC + ĀD + B̄C + B̄D + ABC̄ D̄ Assuming input variables as A, B, C and D, propose a logic circuit using only three logic gates to implement the function. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

चित्र 1(a) में प्रदर्शित द्वि-प्रद्वार जालक्रम के, y-प्राचलों y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 × 10⁻³ S तथा y₁₂ = y₂₁ = 0 के रूप में लक्षण बताए गए हैं । 200 Ω भार के आर-पार वोल्टता का मान ज्ञात कीजिए । (10 अंक)

(b)

चित्र 1(b) में प्रदर्शित संकेत के लिए, संकेत X(t) की संपूर्ण ऊर्जा की गणना कीजिए । y(t) = X(10t – 5) का आरेखण भी कीजिए । (10 अंक)

(c)
(i)

एक 220 V dc समानान्तर क्रम मोटर का, आर्मेचर प्रतिरोध Rₐ = 0·13 Ω, क्षेत्र प्रतिरोध Rf = 250 Ω तथा घूर्णन हान 230 W है । मोटर के 1440 rpm पर पूर्ण भार पर चलते समय लाइन धारा का मान 9·5 A है । निम्नलिखित का मान ज्ञात कीजिए : विकसित (उत्पन्न) यांत्रिक शक्ति

(ii)

निर्गत शक्ति

(iii)

भार (लोड) बल-आघूर्ण

(iv)

पूर्ण भार दक्षता (10 अंक)

(d)

चित्र 1(d) में प्रदर्शित ट्रांजिस्टर परिपथ के लिए, व्युत्क्रम संतृप्त धारा I_S का वह मान ज्ञात कीजिए जो संग्राहक धारा का मान 1 mA कर दे, यदि T = 300 K पर β = 80, V_A = ∞ तथा V_T = 26 mV हो । (10 अंक)

(e)

निम्नानुसार परिभाषित चतुर्वर तार्किक फलन पर विचार कीजिए : F (A, B, C, D) = ĀC + ĀD + B̄C + B̄D + ABC̄ D̄ A, B, C और D को निवेश चर मानकर केवल तीन तार्किक द्वारों का प्रयोग करते हुए इस फलन के कार्यान्वयन के लिए तार्किक परिपथ प्रस्तावित कीजिए । (10 अंक)

Q1 of the 2021 UPSC Mains Electrical Engineering Paper I, as printed
The question as printed in the 2021 Electrical Engineering paper
Evaluate your answer to this question → See all 2021 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 two-port network circuit. An AC voltage source labelled 100 sin (2t) is connected in series with a 2 H inductor. The inductor connects to terminal 1 (top left) of a rectangular 'Two-port Network' block. Terminal 1' (bottom left) of the block connects to the return path of the source. A 200 ohm Load resistor is connected between terminal 2 (top right) and terminal 2' (bottom right) of the two-port network. A 0.01 F capacitor is connected across the bottom terminals 1' and 2'. Current I1 enters terminal 1 and current I2 enters terminal 2. The voltage across the 200 ohm load is to be found.

(b) A graph of X(t) versus t. The horizontal axis is labelled t with tick marks at -3, -2, -1, 0, 1, 2, 3. The vertical axis is labelled X(t) with tick marks at 1 and 2. The waveform is a symmetric trapezoid: it is zero for t < -3, rises linearly from 0 at t = -3 to 2 at t = -2, stays constant at 2 from t = -2 to t = -1, drops linearly from 2 at t = -1 to 1 at t = 0, rises linearly from 1 at t = 0 to 2 at t = 1, stays constant at 2 from t = 1 to t = 2, falls linearly from 2 at t = 2 to 0 at t = 3, and is zero for t > 3.

(d) A bipolar junction transistor circuit. A DC supply Vcc = 2.5 V connects to the collector of an NPN transistor Q1. The base of Q1 connects to one end of a 20 kOhm resistor; the other end of this resistor connects to the collector node. The emitter of Q1 connects to one end of a 1.6 kOhm resistor; the other end of this resistor connects to ground. The quantity asked for is the reverse saturation current IS.

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) Load voltage with the capacitor in the common return Both ports return through the same 0.01 F capacitor, so it carries I₁ + I₂. At ω = 2: jωL = j4 Ω and X_C = 1/(ωC) = 50 Ω. With y₁₂ = y₂₁ = 0, V₁ = I₁/y₁₁ = 303.03 I₁ and V₂ = I₂/y₂₂ = 200 I₂ (I₂ enters port 2 from the load, so the load voltage is −200 I₂). Loop 1: 100 = (303.03 + j4 − j50) I₁ − j50 I₂ Loop 2: 0 = −j50 I₁ + (200 + 200 − j50) I₂ Solving: I₁ = 0.3167 + j0.0463 A and I₂ = −0.01057 + j0.03826 A, |I₂| = 0.0397 A. Load voltage = 200 × 0.0397 ≈ 7.94 V peak, i.e. v(t) ≈ 7.94 sin(2t − 74.6°) V.

(b) Energy of X(t) and the sketch of X(10t − 5) From the figure, X(t) rises linearly 0 → 1 over −3 to −2, is 1 on [−2, −1], is 2 on [−1, 1], is 1 on [1, 2], and falls linearly 1 → 0 over 2 to 3. E = ∫X² dt = 2 × ∫₀¹ t² dt + (1)²(1) + (1)²(1) + (2)²(2) = 2/3 + 2 + 8 = 32/3 ≈ 10.67 units. For y(t) = X(10t − 5), put τ = 10t − 5, so t = (τ + 5)/10: the waveform is compressed by 10 and centred at t = 0.5. The breakpoints τ = −3, −2, −1, 1, 2, 3 map to t = 0.2, 0.3, 0.4, 0.6, 0.7, 0.8. So y rises 0 → 1 on 0.2–0.3, is 1 on 0.3–0.4, is 2 on 0.4–0.6, is 1 on 0.6–0.7 and falls 1 → 0 on 0.7–0.8. Its energy is E/10 = 16/15 ≈ 1.07 units.

Diagram: Signal X(t) and y(t) = X(10t - 5)

(c) Performance of the dc shunt motor I_f = 220/250 = 0.88 A; I_a = 9.5 − 0.88 = 8.62 A. E_b = 220 − 8.62 × 0.13 = 218.88 V. (i) Mechanical power developed = E_b I_a = 218.88 × 8.62 ≈ 1886.7 W. (ii) Power output = 1886.7 − 230 (rotational loss) ≈ 1656.7 W. (iii) Load torque: ω = 2π × 1440/60 = 150.8 rad/s; T = 1656.7/150.8 ≈ 10.99 N·m. (iv) Efficiency: input = 220 × 9.5 = 2090 W; η = 1656.7/2090 ≈ 79.3%.

(d) Reverse saturation current of the PNP transistor The transistor is a PNP: emitter at V_CC = 2.5 V, base returned to the collector node through 20 kΩ, and collector to ground through 1.6 kΩ. I_C = 1 mA, so I_B = 1/80 = 12.5 μA. The 20 kΩ carries I_B into the collector node, so the 1.6 kΩ carries I_C + I_B = 1.0125 mA. V_C = 1.0125 mA × 1.6 kΩ = 1.62 V; V_B = 1.62 + 12.5 μA × 20 kΩ = 1.87 V. V_EB = 2.5 − 1.87 = 0.63 V (V_EC = 0.88 V, so the device is active). I_C = I_S exp(V_EB/V_T) gives I_S = 10⁻³ × exp(−0.63/0.026) = 10⁻³/3.34 × 10¹⁰. I_S ≈ 3.0 × 10⁻¹⁴ A.

(e) Three-gate realisation F = Ā(C + D) + B̄(C + D) + AB·C̄D̄ = (Ā + B̄)(C + D) + AB·(C + D)′ = (AB)′(C + D) + (AB)·(C + D)′ F = (A·B) ⊕ (C + D) Circuit: one AND gate (inputs A, B → X), one OR gate (inputs C, D → S), one XOR gate (inputs X, S → F). Check: A = B = 1, C = D = 0 gives X = 1, S = 0, F = 1, as in the original term ABC̄D̄.

How this answer will be evaluated

What earns full marks: Rigorous method, correct units, all sub-parts fully solved with clear diagrams.

Evaluation rubric

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

  1. (a) Load voltage using the y-parameter model, with the capacitor in the common return of both ports. 10 marks

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

    Must cover

    • Recognise that the 0.01 F capacitor carries I₁ + I₂, with X_C = 50 Ω and jωL = j4 Ω at ω = 2.
    • Write the two loop equations using V₁ = I₁/y₁₁ and V₂ = I₂/y₂₂ and solve them.
    • State the load voltage as about 7.94 V peak with phase about −74.6°.

    Loses marks

    • Ignoring the capacitor common to both ports (for example, declaring V₂ = 0).
    • Treating y-parameters as z-parameters or dropping the inductor.

    Earns more

    • Correct phasor values for ω, jωL and X_C.
    • Correct decoupling of the ports because y₁₂ = y₂₁ = 0.

    Extra mark

    • Time-domain form v(t) = 7.94 sin(2t − 74.6°) V.
  2. (b) Energy of X(t) from the figure and the sketch of y(t) = X(10t − 5). 10 marks

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

    Must cover

    • Read the waveform correctly and integrate X² piecewise to get E = 32/3.
    • Apply y(t) = X(10t − 5): compression by 10 and centring at t = 0.5, with breakpoints 0.2, 0.3, 0.4, 0.6, 0.7, 0.8.
    • Sketch with levels 0, 1, 2, 1, 0 and labelled axes.

    Loses marks

    • Claiming energy is unchanged by time scaling.
    • A sketch with wrong width, shift or amplitudes (for example, shifting before scaling).

    Earns more

    • Energy scaling E_y = E_x/10 = 16/15 stated.
    • Clear labelling of breakpoints and amplitudes on the sketch.

    Extra mark

    • Direct integration of y² to confirm E_y = E_x/10.
  3. (c) Mechanical power developed, output power, load torque and efficiency of the dc shunt motor. 10 marks

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

    Must cover

    • Find I_f = 0.88 A and I_a = 8.62 A.
    • Developed power E_b I_a = 1886.7 W and output = 1886.7 − 230 = 1656.7 W.
    • Torque = P_out/ω = 10.99 N·m and efficiency = 1656.7/2090 = 79.3%.

    Loses marks

    • Using the line current instead of the armature current for E_b I_a.
    • Leaving out the rotational loss from the output or taking efficiency on developed power.

    Earns more

    • Correct E_b = 220 − 8.62 × 0.13 = 218.88 V.
    • Correct ω = 48π rad/s with units.

    Extra mark

    • Loss breakdown: field 193.6 W, armature 9.66 W, rotational 230 W.
  4. (d) Reverse saturation current I_S of the PNP transistor in Figure 1(d). 10 marks

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

    Must cover

    • Identify the transistor as PNP, with the base returned to the collector node through 20 kΩ.
    • Find I_B = 12.5 μA, V_C = 1.62 V, V_B = 1.87 V and V_EB = 0.63 V (1.6 kΩ carries I_C + I_B).
    • Use I_C = I_S exp(V_EB/V_T) to get I_S ≈ 3 × 10⁻¹⁴ A.

    Loses marks

    • Treating the device as NPN.
    • Leaving I_B out of the drop across the 1.6 kΩ resistor.

    Earns more

    • Correct I_B = I_C/β.
    • Correct inversion I_S = I_C exp(−V_EB/V_T) with units.

    Extra mark

    • Check of the active region (V_EC = 0.88 V).
  5. (e) A three-gate logic circuit for F(A, B, C, D). 10 marks

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

    Must cover

    • Simplify F to (AB)′(C + D) + AB(C + D)′.
    • Recognise F = (A·B) ⊕ (C + D).
    • Implement with one AND, one OR and one XOR gate and verify.

    Loses marks

    • Using more than three gates.
    • A wrong simplification such as F = (A + B)′(C + D).

    Earns more

    • Step-by-step Boolean simplification shown.
    • Inputs, intermediate outputs and F labelled.

    Extra mark

    • Truth-table check for a few input combinations.
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.

All UPSC directive words, compared →

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Electrical Engineering 2021 Paper I

Write this answer and get it marked in about 20 seconds

Free account: 5 AI evaluations a month, marked against the rubric above.

Start free

Already have an account? Log in