Chat with us
Electrical Engineering 2023 Paper I 50 marks Compulsory Solve

Paper I — Q1

(a) Obtain Norton equivalent circuit at terminals ab of the coupled circuit shown in the figure. Using it, find out the current…

(a)

Obtain Norton equivalent circuit at terminals ab of the coupled circuit shown in the figure. Using it, find out the current passing through 5 Ω resistor connected between the terminals ab. 10 marks

(b)

Obtain the Laplace transform of the following periodic waveforms : (10 marks) (i) (ii)

(c)

A 3-phase, 50 Hz, star-connected cage-type induction motor has standstill input impedance of (1·0 + j 3·0) Ω per phase. The motor is connected through a cable from 400 V, 3-phase balanced supply so that the blocked rotor voltage at its terminal is dropped by 20% from the supplied voltage. The motor is to be started through a DOL starter from the same supply and cable as above. Find : (i) the cable impedance per phase, (ii) the motor starting current, (iii) input power factor at the time of starting. (Assume negligible stator impedance of the motor and cable R/X ratio of 3 : 1 at 50 Hz supply. Also ignore magnetizing current and core losses.) 10 marks

(d)

Calculate the lower corner frequency for the circuit shown below. Take transistor parameters as : β = 100, V_BE = 0·7 V and V_A = ∞. (10 marks) V_CC = 12 V, R_1 = 10 kΩ, R_S = 0·5 kΩ, C_C = 0·1 μF, R_2 = 1·5 kΩ, R_C = 1 kΩ, R_E = 0·1 kΩ

(e)

A metal bar slides over a pair of conducting rails in a uniform magnetic field B⃗ = a⃗_z B_0 Wb/m² with a constant velocity u⃗ m/s as shown below in the figure. A resistance 'R' Ω is connected between terminals 1 and 2. Prove that this system upholds the principle of conservation of energy. Neglect the electrical resistance of the metal bar and the pair of conducting rails, and the mechanical friction of this ideal system. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

चित्र में दर्शाए गए युग्मित परिपथ का ab अन्तस्थों पर नॉर्टन समतुल्य परिपथ निकालिए । उसका उपयोग करते हुए, ab अन्तस्थों के मध्य जुड़े 5 Ω के प्रतिरोधक से प्रवाहित होने वाली धारा ज्ञात कीजिए । (10 अंक)

(b)

निम्नलिखित आवर्ती तरंगरूपों का लाप्लास रूपांतरण निकालिए : (10 अंक) (i) (ii)

(c)

एक त्रि-कला, 50 Hz तारा-संयोजित पंजर प्रेरण मोटर की विराम निवेश प्रतिबाधा (1·0 + j 3·0) Ω प्रति कला है । यह मोटर 400 V, त्रि-कला संतुलित विद्युत प्रदाय से केबल के माध्यम से जोड़ी गई है जिससे इसके सिरे पर अवरोधित घूर्णक बोल्टता संभरण बोल्टता से 20% कम हो । इस मोटर को उपयुक्त संभरण बोल्टता एवं केबल द्वारा एक DOL स्टार्टर के माध्यम से शुरु किया जाना है । ज्ञात कीजिए : (i) केबल की प्रतिबाधा प्रति कला, (ii) मोटर की आरंभिक धारा, (iii) मोटर के शुरू होते समय निवेश शक्ति गुणांक । (मोटर की स्टेटर प्रतिबाधा को नगण्य एवं 50 Hz विद्युत प्रदाय पर केबल का R/X अनुपात 3 : 1 मानिए । साथ ही चुंबकीय धारा एवं क्रोड हानियों को भी अनदेखा कीजिए ।) (10 अंक)

(d)

नीचे दर्शाए गए परिपथ के लिए निम्न कोणा आवृत्ति निकालिए । ट्रांजिस्टर के प्राचल निम्नानुसार हैं : β = 100, V_BE = 0·7 V एवं V_A = ∞ । (10 अंक)

(e)

चित्र में दर्शाए अनुसार एकसमान चुंबकीय क्षेत्र B⃗ = a⃗_z B_0 Wb/m² में एक धातु पट्टी एक सुचालक रेल के जोड़े के ऊपर अचर बेग u⃗ m/s से फिसलती है । अंतस्थ 1 और 2 के मध्य एक प्रतिरोध 'R' Ω जुड़ा है । सिद्ध कीजिए कि यह तंत्र ऊर्जा संरक्षण के सिद्धांत का अनुमोदन करता है । धातु पट्टी एवं सुचालक रेल के जोड़े के विद्युतीय प्रतिरोध तथा इस आदर्श तंत्र के यांत्रिक घर्षण को अनदेखा कीजिए । (10 अंक)

Q1 of the 2023 UPSC Mains Electrical Engineering Paper I, 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 figures printed on the question paper

Cut from the original 2023 Electrical Engineering paper, exactly as the candidates in the hall saw them.

Part (b)
Figure for part (b) of this question
Part (b)
Figure for part (b) of this question
Part (d)
Figure for part (d) of this question
Part (e)
Figure for part (e) 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.

(a) A circuit diagram featuring two voltage sources and coupled inductors. On the left, a voltage source of 10∠0° V is connected in series with a 4 Ω resistor. This connects to a node marked with a dot, which is the start of an inductor with impedance j8 Ω. On the right, a voltage source of 10∠90° V is connected in series with a 2 Ω resistor. This connects to a node marked with a dot, which is the start of an inductor with impedance j4 Ω. The two inductors are magnetically coupled with a mutual impedance of j5 Ω, indicated by an arc connecting the two dotted nodes. The other ends of the j8 Ω and j4 Ω inductors are connected together at a central node. Terminal 'a' is connected to this central node. Terminal 'b' is connected to the common bottom wire that links the negative terminals of both voltage sources.

(b) A circuit diagram showing four switches labeled SW1, SW2, SW3, and SW4. Each switch is connected in series with a 5 V voltage source (represented by a resistor symbol with a 5 V label). The other ends of the switches are connected to a common ground. The voltage sources are connected to the input of a block labeled 'Logic Circuit'. The output of the logic circuit is labeled X. Below the diagram, a caption reads: (SW1, SW2, SW3, SW4 : Switch 1, Switch 2, Switch 3, Switch 4).

(c) Circuit diagram showing a two-port network. On the left, a voltage source labeled V_S1 with positive terminal up is connected in series with the input port of a rectangular box labeled 'Network'. The current entering the top terminal of the network is labeled I_1. On the right, a voltage source labeled V_S2 with positive terminal up is connected in series with the output port of the network. The current entering the top terminal of the network from the right side is labeled I_2. The bottom terminals of both sources and the network are connected by a common wire.

Below the circuit is a table with 5 rows and 5 columns. Header row: (empty), V_S1 Volts, V_S2 Volts, I_1 Amp, I_2 Amp Row 1: Experiment 1, 100, 50, 5, -30 Row 2: Experiment 2, 50, 100, -20, -5 Row 3: Experiment 3, 25, 0, -, - Row 4: Experiment 4, -, -, 5, 0

(d) A BJT common-emitter amplifier circuit. A DC voltage source V_CC = 12 V is connected to the top rail. A voltage divider bias network consists of resistor R_1 = 10 kΩ connected between V_CC and the base, and resistor R_2 = 1.5 kΩ connected between the base and ground. The collector is connected to V_CC through a resistor R_C = 1 kΩ. The emitter is connected to ground through a resistor R_E = 0.1 kΩ. The output voltage v_o is taken at the collector. The input signal v_i is connected in series with a source resistance R_S = 0.5 kΩ and a coupling capacitor C_C = 0.1 μF, which connects to the base of the transistor. The transistor is an NPN type.

(e) A 2D diagram showing a metal bar sliding on a pair of horizontal conducting rails. The setup is in the X-Y plane. The rails are connected on the left side to a resistor labeled 'R'. The terminals of the resistor are labeled '1' (top) and '2' (bottom). A vertical metal bar bridges the two rails and is shown moving to the right with a velocity vector labeled 'u'. The distance between the rails is labeled 'h meters'. The region between the rails is filled with circles containing dots, representing a uniform magnetic field directed out of the page, labeled 'Uniform magnetic field'. The X-axis points to the right, and the Y-axis points up.

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) Norton equivalent at ab, and current in the 5 Ω resistor

Both coil dots are at the outer ends. With ab open, the same loop current I flows through the two branches, entering L2 at its undotted end, so the coupling opposes.

Loop: (4 + 2 + j8 + j4 − 2·j5) I = 10∠0° − 10∠90°, so (6 + j2) I = 10 − j10, giving I = 1 − j2 A.

Open-circuit voltage: V_ab = 10 − (4 + j8 − j5)(1 − j2) = 10 − (4 + j3)(1 − j2) = 10 − (10 − j5) = j5 V = 5∠90° V. The right-hand branch gives the same value.

Impedance with both sources shorted: drive a current I_t into a and let it split as I1 (left) and I2 (right); both enter dotted ends, so the coupling aids. Then V_a = (4 + j8) I1 + j5 I2 = (2 + j4) I2 + j5 I1. The second equality gives (4 + j3) I1 = (2 − j) I2, so I1 = (0.2 − j0.4) I2. Then V_a = (4 + j5) I2 and I_t = (1.2 − j0.4) I2, so Z_N = (4 + j5)/(1.2 − j0.4) = 1.75 + j4.75 Ω.

Norton current: I_N = V_oc / Z_N = j5/(1.75 + j4.75) = 0.927 + j0.341 A = 0.988∠20.2° A, in parallel with Z_N.

Load current: I_5 = I_N · Z_N/(Z_N + 5) = j5/(6.75 + j4.75) = 0.349 + j0.495 A = 0.606∠54.9° A.

Diagram: Norton equivalent at terminals ab with the 5 ohm load connected

(b) Laplace transforms of the periodic waveforms

For a periodic function, F(s) = F₁(s)/(1 − e^(−sT)), where F₁ is the transform of the first period.

(i) Sawtooth. Period T = a and f(t) = t/a for 0 < t < a. Then F₁(s) = (1/a)∫₀^a t e^(−st) dt = [1 − (1 + as)e^(−as)]/(a s²). Hence F(s) = [1 − (1 + as)e^(−as)] / [a s² (1 − e^(−as))] = 1/(a s²) − 1/[s(e^(as) − 1)].

(ii) Trapezoid. Period T = 4a: the wave rises with slope 1/a over 0 to a, is flat at 1 from a to 3a, and falls with slope −1/a from 3a to 4a. One period is f₁(t) = (1/a)[t u(t) − (t − a)u(t − a) − (t − 3a)u(t − 3a) + (t − 4a)u(t − 4a)]. So F₁(s) = (1 − e^(−as) − e^(−3as) + e^(−4as))/(a s²) = (1 − e^(−as))(1 − e^(−3as))/(a s²), and F(s) = (1 − e^(−as))(1 − e^(−3as)) / [a s² (1 − e^(−4as))].

(c) Star-connected motor on a cable, direct-on-line start

Per-phase supply V = 400/√3 = 230.9 V. Motor impedance Z_m = 1 + j3 Ω, |Z_m| = 3.162 Ω. At standstill (which is the starting condition) the terminal voltage is 0.8 V, so |Z_m + Z_c| = |Z_m|/0.8 = 3.953 Ω.

(i) Cable impedance. With R_c = r and X_c = r/3: (1 + r)² + (3 + r/3)² = 15.625, so 1.111r² + 4r − 5.625 = 0, giving r = 1.081 Ω and X_c = 0.360 Ω. Z_c = 1.081 + j0.360 Ω = 1.14∠18.4° Ω.

(ii) Starting current. I_st = 230.9/3.953 = 58.4 A per phase (equal to the line current in a star connection). Check: 58.4 × 3.162 = 184.8 V = 0.8 × 230.9 V.

(iii) Power factor. At the motor terminals, cos φ = 1/3.162 = 0.316 lagging. At the supply end, including the cable, cos φ = (1 + 1.081)/3.953 = 0.527 lagging.

(d) Lower corner frequency

Only the coupling capacitor C_C sets the lower corner, since the emitter resistor is not bypassed.

Bias: V_th = 12 × 1.5/11.5 = 1.565 V and R_th = 10 ∥ 1.5 = 1.304 kΩ. Then I_B = (1.565 − 0.7)/(1.304 + 101 × 0.1) = 75.9 µA, I_C = 7.59 mA and I_E = 7.66 mA. V_CE = 12 − 7.59 × 1 − 7.66 × 0.1 = 3.65 V, so the transistor is in the active region.

Small signal (V_T = 25 mV): g_m = 7.59/25 = 0.303 S and r_π = β/g_m = 330 Ω.

Input resistance seen at the base: r_π + (β + 1)R_E = 0.33 + 10.1 = 10.43 kΩ. With the bias resistors: R_in = 10 ∥ 1.5 ∥ 10.43 = 1.159 kΩ.

f_L = 1/[2π (R_S + R_in) C_C] = 1/[2π × 1.659 kΩ × 0.1 µF] ≈ 959 Hz, about 0.96 kHz.

(e) Sliding bar and conservation of energy

The bar of length h moves at speed u across field B₀. Motional EMF: e = B₀ h u, with u × B pointing along −y, so current flows down the bar, left along the lower rail, up through R (from terminal 2 to 1) and back along the upper rail. Current I = B₀ h u / R.

Force on the bar: F = I h B₀ = B₀² h² u / R, directed along −x, opposing the motion (Lenz's law). To keep u constant, an external agent must apply an equal force along +x.

Mechanical power supplied: P_mech = F u = B₀² h² u² / R.

Electrical power dissipated in R: P_el = I² R = (B₀ h u/R)² R = B₀² h² u² / R.

P_mech = P_el, and with no friction or resistance elsewhere, all supplied mechanical energy appears as heat in R, which upholds conservation of energy.

Diagram: Bar sliding on rails in a field out of the page: motional EMF, current through R and opposing force

How this answer will be evaluated

What earns full marks: Complete circuit models, correct derivations, all sub-parts solved with proper units and phasor diagrams where applicable.

Evaluation rubric

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

  1. (a) Obtain the Norton equivalent at ab of the coupled circuit and find the current in the 5 ohm resistor. 10 marks

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

    Must cover

    • Open-circuit voltage j5 V using correct coupling sign (loop current 1 - j2 A)
    • Norton impedance 1.75 + j4.75 ohm with both sources shorted
    • Norton current 0.927 + j0.341 A
    • Load current 0.606 angle 54.9 degrees A

    Loses marks

    • Sign error in the mutual term or no Norton circuit.

    Earns more

    • Marks for the correct mutual-coupling treatment and each step with units.

    Extra mark

    • Cross-check of load current by direct mesh analysis.
  2. (b) Obtain the Laplace transform of the two periodic waveforms. 10 marks

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

    Must cover

    • Periods T = a and T = 4a correctly identified
    • F1(s) by integration or step functions
    • Division by (1 - e^(-sT))
    • Final expressions for (i) and (ii)

    Loses marks

    • Missing the division by (1 - e^(-sT)).

    Earns more

    • Marks for each waveform with the first-period transform.

    Extra mark

    • Unit-step method for the trapezoid.
  3. (c) Find the cable impedance, starting current and power factor for the DOL start. 10 marks

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

    Must cover

    • Terminal voltage 0.8 of supply gives |Zm + Zc| = 3.953 ohm
    • Cable Zc = 1.081 + j0.360 ohm from R/X = 3
    • Starting current 58.4 A
    • Power factor 0.316 at motor, 0.527 at supply

    Loses marks

    • Mixing line and phase voltage.

    Earns more

    • Marks for per-phase equivalent circuit and each result.

    Extra mark

    • Stating both power factors.
  4. (d) Calculate the lower corner frequency of the amplifier. 10 marks

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

    Must cover

    • Bias: IB about 75.9 uA, IC about 7.59 mA, VCE about 3.65 V
    • gm about 0.303 S, r-pi about 330 ohm
    • Rin with unbypassed RE and bias resistors about 1.16 kohm
    • fL = 1/(2 pi (Rs + Rin) Cc), about 0.96 kHz

    Loses marks

    • Ignoring the source resistance or the emitter resistor.

    Earns more

    • Marks for bias, small-signal model and correct time constant including Rs.

    Extra mark

    • Stating the active-region check.
  5. (e) Prove that the sliding bar system upholds conservation of energy. 10 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • EMF = B0 h u and current = B0 h u / R with direction
    • Force B0^2 h^2 u / R opposing motion
    • Mechanical power F u
    • Electrical power I^2 R equal to it

    Loses marks

    • Omitting the opposing force or the power comparison.

    Earns more

    • Marks for EMF, force, both powers and the equality.

    Extra mark

    • Lenz's law and direction of current.
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 2023 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