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Electrical Engineering 2023 Paper I 50 marks Solve

Paper I — Q6

(a) A 200 V, 1100 rpm, dc shunt motor takes 1·5 A and runs at 1150 rpm under no load condition at rated voltage. Its armature…

(a)
(i)

A 200 V, 1100 rpm, dc shunt motor takes 1·5 A and runs at 1150 rpm under no load condition at rated voltage. Its armature resistance including brushes is 0·5 Ω. While running under full load, its field circuit gets open circuited due to fault and the motor takes 5 times of its rated input current to deliver rated torque. Find : the full load torque of the machine, and

(ii)

the speed of the motor under field fault condition. Assume no armature copper losses at no load and no armature reaction. 20 marks

(b)

An absorber material of relative permeability and relative permittivity of εᵣ = μᵣ = 6 – j6 is coated on a perfectly conducting sheet and this combination is placed in free space as shown in the figure given below. A 500 MHz wave is incident on it normally from free space. Calculate the thickness of the absorber required to attenuate the reflected wave by 30 dB. 20 marks

(c)

A 120 V, 1000 rpm, 350 W separately excited dc motor is supplied via half-controlled single-phase bridge rectifier for speed control purpose. The supply voltage to the rectifier is 200 V, 50 Hz ac and to obtain the desired speed, the triggering angle is set at 105° at one instant. The armature current is discontinuous with an average value of 2 A and it continues up to 30° beyond voltage zero. The motor armature resistance is 1·5 Ω. Determine the operating speed of the motor. (Assume constant flux operation and no armature reaction) 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

एक 200 V, 1100 rpm, dc समानांतर क्रम मोटर निर्धारित वोल्टता पर भार रहित अवस्था में 1150 rpm पर चलती है और 1·5 A लेती है। इसका ब्रश सहित आर्मेचर प्रतिरोध 0·5 Ω है। जब यह पूर्ण भार पर चल रही होती है, तो दोष के कारण इसकी क्षेत्र कुंडली खुले परिपथ में हो जाती है और मोटर निर्धारित बलाघूर्ण प्रदान करने के लिए, निर्धारित निवेश धारा से 5 गुना धारा लेती है। ज्ञात कीजिए : यंत्र का पूर्ण भार बलाघूर्ण, और

(ii)

क्षेत्र दोष की अवस्था में मोटर की गति। मान लीजिए कि भार रहित अवस्था में आर्मेचर कॉपर हास नहीं हैं और आर्मेचर रिएक्शन भी नहीं है। (20 अंक)

(b)

एक शोषक पदार्थ जिसके सापेक्ष पारगम्यता और सापेक्ष परावैद्युतांक के मान εr = μr = 6 – j6 हैं, को एक आदर्श सुचालक पतर के ऊपर लेपित किया गया है और यह संयोजन, मुक्त अंतराल में रखा गया है जैसा कि चित्र में प्रदर्शित है। मुक्त अंतराल से एक 500 MHz की तरंग इस पर लम्बवत आपतित है। परावर्तित तरंग के 30 dB क्षीणन के लिए शोषक की वांछित मोटाई की गणना कीजिए। (20 अंक)

(c)

एक 120 V, 1000 rpm, 350 W अलग से उत्तेजित dc मोटर, गति नियंत्रण के उद्देश्य से अर्ध-नियंत्रित एकल-कला ब्रिज दिष्टकारी द्वारा संभारित है। दिष्टकारी के लिए संभारण वोल्टता 200 V, 50 Hz ac है और वांछित गति प्राप्त करने के लिए एक समय उत्त्प्रेरण (ट्रिगर) कोण 105° समायोजित किया गया है। आर्मेचर धारा औसत मान 2 A के साथ असतत है और यह वोल्टता शून्य के 30° परे तक निरंतर रहती है। मोटर आर्मेचर का प्रतिरोध 1·5 Ω है। मोटर की परिचालन गति ज्ञात कीजिए। (मान लीजिए कि स्थिर अभिवाह परिचालन है और आर्मेचर प्रतिक्रिया नहीं है) (10 अंक)

Q6 of the 2023 UPSC Mains Electrical Engineering Paper I, as printed
The question as printed in the 2023 Electrical Engineering paper
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Cut from the original 2023 Electrical Engineering paper, exactly as the candidates in the hall saw it.

Part (b)
Figure for part (b) of this question

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 schematic diagram showing a rectangular block labeled 'Absorber material' positioned between a region labeled 'Free Space' on the left and a vertical hatched line on the right labeled 'Conducting sheet'. A horizontal arrow labeled 'Incident wave' points from the left towards the absorber material. A double-headed arrow below the block indicates the thickness of the absorber material.

(c) Circuit diagram: A two-port network (labeled 'Network' or 'तंत्र') is connected between two voltage sources. On the left, a voltage source V_S1 is connected to the input port. The positive terminal is at the top. Current I_1 flows into the top terminal of the network. On the right, a voltage source V_S2 is connected to the output port. The positive terminal is at the top. Current I_2 flows into the top terminal of the network (indicated by an arrow pointing left). The bottom terminals of both sources and the network are connected together. Below the circuit is a table with 5 rows and 5 columns. The header row contains: (blank), V_S1 Volts, V_S2 Volts, I_1 Amp, I_2 Amp. Row 1 (Experiment 1 / प्रयोग 1): 100, 50, 5, -30. Row 2 (Experiment 2 / प्रयोग 2): 50, 100, -20, -5. Row 3 (Experiment 3 / प्रयोग 3): 25, 0, -, -. Row 4 (Experiment 4 / प्रयोग 4): -, -, 5, 0.

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) Shunt motor: full-load torque and speed on field failure

At no load the armature copper loss is neglected, so the 1.5 A is the field current and the back EMF equals the terminal voltage: E_b0 = 200 V at 1150 rpm. With constant flux, E_b ∝ N.

(i) Full-load torque. At 1100 rpm: E_b = 200 × 1100/1150 = 191.3 V, so I_a = (200 − 191.3)/0.5 = 17.39 A and the rated input current is 17.39 + 1.5 = 18.89 A. Developed power = E_b I_a = 191.3 × 17.39 = 3327 W and ω = 2π × 1100/60 = 115.2 rad/s, so T = 3327/115.2 = 28.9 N·m.

(ii) Speed after the field opens. The input current is now all armature current: I_a′ = 5 × 18.89 = 94.46 A. The torque is unchanged, T ∝ φ I_a, so φ′/φ = 17.39/94.46 = 0.184 (residual flux). The new back EMF is E_b′ = 200 − 94.46 × 0.5 = 152.8 V. Since N ∝ E_b/φ: N′ = 1100 × (152.8/191.3) × (1/0.184) = 1100 × 0.7986 × 5.431 ≈ 4770 rpm.

This is a runaway speed, about 4.3 times rated, which is why shunt motors need field-failure protection.

(b) Thickness of the absorber layer

Since ε_r = μ_r, the intrinsic impedance of the absorber is η = η₀√(μ_r/ε_r) = η₀, which matches free space, so nothing is reflected at the air-absorber interface. The only reflected wave is the one that travels through the layer, is reflected by the perfect conductor (Γ = −1) and returns, so its amplitude is |Γ| = e^(−2αd).

Propagation constant: γ = jω√(μ₀ε₀ μ_rε_r) = jk₀(6 − j6) = 6k₀ + j6k₀, where k₀ = 2π × 500 × 10⁶/(3 × 10⁸) = 10.47 rad/m. So α = 6k₀ = 62.83 Np/m.

30 dB reduction means 20 log₁₀ e^(2αd) = 30, so e^(2αd) = 10^1.5 = 31.6 and 2αd = 3.454. Hence d = 3.454/(2 × 62.83) = 0.0275 m, about 27.5 mm.

Diagram: Absorber layer of thickness d on a perfectly conducting sheet with incident wave, reflected wave and amplitude decay through the layer

(c) Separately excited motor on a half-controlled bridge

Take the supply as 200 V rms, so V_m = 200√2 = 282.8 V, with α = 105° and extinction angle β = 180° + 30° = 210°. The current flows from 105° to 210° and is zero from 210° to the next firing at 285°, which is 75° of each 180° period.

During 105° to 180° the motor sees the supply; from 180° to 210° the freewheeling diode holds the armature voltage at zero; and for the remaining 75° the armature voltage equals the back EMF E.

Average terminal voltage: V₀ = (V_m/π)(1 + cos α) + E × 75/180 = 90.03 × 0.7412 + 0.4167E = 66.73 + 0.4167E.

Also V₀ = E + I_a R_a = E + 2 × 1.5 = E + 3. So 0.5833E = 63.73, E = 109.3 V.

Rated back EMF: take the rated armature current as 350/120 = 2.92 A, so E_rated = 120 − 2.92 × 1.5 = 115.6 V at 1000 rpm.

Speed N = 1000 × 109.3/115.6 ≈ 945 rpm (constant flux, so N ∝ E).

Diagram: Half-controlled bridge feeding a dc motor: supply voltage, armature voltage with supply, freewheeling and back EMF intervals, and discontinu

How this answer will be evaluated

Framework: Electrical Engineering, Paper 1

What earns full marks: Correct method with all steps shown, proper units, and clear interpretation of results.

Evaluation rubric

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

  1. (a) Find the full-load torque and the speed of the shunt motor after field failure. 20 marks

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

    Must cover

    • No-load: Eb0 = 200 V at 1150 rpm; 1.5 A is field current
    • Full load: Eb = 191.3 V, Ia = 17.39 A, input 18.89 A, T = 28.9 N m
    • Field failure: Ia' = 94.46 A, flux ratio 0.184, Eb' = 152.8 V
    • Speed about 4770 rpm

    Loses marks

    • Keeping the flux constant after the field opens.

    Earns more

    • Marks for the EMF, torque and speed steps.

    Extra mark

    • Comment on runaway and protection.
  2. (b) Calculate the absorber thickness for 30 dB attenuation of the reflected wave. 20 marks

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

    Must cover

    • eta = eta0 because mu_r = eps_r, so no interface reflection
    • Gamma at the conductor = -1; |reflected| = exp(-2 alpha d)
    • gamma = j k0 (6 - j6), k0 = 10.47 rad/m, alpha = 62.83 Np/m
    • d = 3.454/(2 alpha) = 27.5 mm

    Loses marks

    • Treating 30 dB as one-way attenuation.

    Earns more

    • Marks for the matched impedance, round-trip loss and the thickness.

    Extra mark

    • Noting phase constant beta = alpha.
  3. (c) Determine the operating speed of the motor on the half-controlled bridge. 10 marks

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

    Must cover

    • alpha = 105 degrees, beta = 210 degrees, zero-current interval 75 degrees
    • V0 = (Vm/pi)(1 + cos alpha) + E x 75/180
    • V0 = E + Ia Ra gives E = 109.3 V
    • Speed about 945 rpm using rated EMF 115.6 V

    Loses marks

    • Using the continuous-conduction formula.

    Earns more

    • Marks for the average voltage with the zero-current term and the speed ratio.

    Extra mark

    • Stating the rated-current assumption.
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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.

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