Q5
(a) The diagram of Master-Slave S-R flip-flop and the waveform applied to the Master flip-flop is shown in Figure 5(a). Draw the waveform that appears at the output of Slave flip-flop. 10 marks (b) An ideal lossless λ/4 extension line of Z₀ = 60 Ω is terminated with a load resistance of 60 Ω. Find the value of Z_in. 10 marks (c) The power supplied to a 3-phase induction motor is 40 kW and the corresponding stator losses are 1·5 kW. Calculate the net (shaft) mechanical power developed and the rotor Cu loss, when the slip is 0·04 pu. What will be the net power developed if the speed of the above motor is reduced to 40% of the synchronous speed by means of external rotor resistance, assuming the torque and stator losses remain unaltered ? Friction and windage losses may be assumed to be 0·8 kW. 10 marks (d) An audio frequency signal 10 sin (2π × 500 t) is used to amplitude modulate a carrier of 50 sin (2π × 10⁵ t). Determine : (i) the modulation index (ii) the amplitude of each sideband frequency (iii) the bandwidth required (iv) total power delivered to the load of 500 Ω (v) and draw the frequency spectrum. 10 marks (e) For a two-port network, the currents I₁ and I₂ are as given below : I₁ = 2V₁ – V₂, I₂ = – V₁ + 2V₂. Find the transmission and hybrid parameters of the network. 10 marks
हिंदी में प्रश्न पढ़ें
(a) एक मास्टर-स्लेव S-R फ्लिप-फ्लॉप व उसके मास्टर फ्लिप-फ्लॉप पर अनुप्रयुक्त तरंग रूप का आरेख चित्र 5(a) में दर्शाया गया है । स्लेव फ्लिप-फ्लॉप के निर्गम पर प्रकट होने वाले तरंग रूप का आरेखण कीजिए । 10 अंक (b) Z₀ = 60 Ω वाली एक आदर्श क्षयहीन λ/4 विस्तार लाइन एक 60 Ω के भार प्रतिरोध के साथ अंतस्थ होती है । Z_in का मान ज्ञात कीजिए । 10 अंक (c) एक त्रि-कला प्रेरण मोटर को 40 kW शक्ति की आपूर्ति की जा रही है, तदनुसार स्टेटर में 1·5 kW शक्ति हानि होती है । जब सर्पण 0·04 pu हो, तो रोटर में ताम्र हानि व उत्पन्न शुद्ध (शैफ्ट) यांत्रिक शक्ति की गणना कीजिए । यदि उपर्युक्त मोटर की गति को बाह्य रोटर प्रतिरोध की सहायता से तुल्यकालिक गति के 40% तक कम कर दिया जाए, तो बल-आघूर्ण व स्टेटर हानि को अपरिवर्तित मान कर, उत्पन्न शुद्ध शक्ति का मान क्या होगा ? यांत्रिक घर्षण हानि व वायु घर्षण हानि को 0·8 kW मान लीजिए । 10 अंक (d) एक वाहक 50 sin (2π × 10⁵ t) का एक श्रव्य आवृत्ति संकेत 10 sin (2π × 500 t) के द्वारा आयाम मॉडुलन किया जाता है । निर्धारित कीजिए : (i) मॉडुलन सूचकांक (ii) प्रत्येक पार्श्व बैंड आवृत्ति का आयाम (iii) आवश्यक बैंड विस्तार (iv) 500 Ω भार को प्रदत्त पूर्ण शक्ति (v) तथा आवृत्ति स्पेक्ट्रम का आरेखण कीजिए । 10 अंक (e) एक द्वि-प्रद्वार जालक्रम के लिए धारा I₁ व I₂ के मान निम्न प्रकार हैं : I₁ = 2V₁ – V₂, I₂ = – V₁ + 2V₂ । जालक्रम के संचरण व संकर प्राचल ज्ञात कीजिए । 10 अंक
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Approach
Solve all five sub-parts systematically, allocating approximately 20% time to each part given equal 10-mark weighting. For (a), clearly show the master-slave timing relationship with proper edge triggering; for (b)-(e), present formulas first, then substitution, then final numerical answers with units. Use separate sections for each sub-part with clear labels.
Key points expected
- (a) Master-Slave S-R flip-flop: Correct identification of level-triggered master and edge-triggered slave operation; output changes only on falling edge of clock; proper handling of S=R=1 forbidden state
- (b) Quarter-wave transformer: Application of Z_in = Z₀²/Z_L formula; recognition that matched load (Z_L = Z₀) yields Z_in = Z₀ = 60 Ω regardless of line length
- (c) Induction motor power flow: Air-gap power P_g = P_in - P_stator = 38.5 kW; rotor Cu loss = sP_g = 1.54 kW; P_mech = (1-s)P_g = 36.96 kW; shaft power = P_mech - P_fw = 36.16 kW; at new slip s' = 0.6, maintaining same torque implies same P_g, hence new rotor Cu loss = 23.1 kW and new shaft power calculation
- (d) AM modulation: m = A_m/A_c = 0.2; sideband amplitude = mA_c/2 = 5 V; bandwidth = 2f_m = 1 kHz; total power = P_c(1 + m²/2) with P_c = A_c²/(2R) = 2.5 W giving 2.55 W; frequency spectrum showing carrier at 100 kHz and sidebands at 99.5 kHz and 100.5 kHz
- (e) Two-port parameters: From given Y-parameters (y11=2, y12=-1, y21=-1, y22=2 S), derive ABCD parameters [A=-2, B=-1Ω, C=-3S, D=-2] and h-parameters [h11=0.5Ω, h12=0.5, h21=-0.5, h22=1.5S] using standard conversion formulas
Evaluation rubric
| Dimension | Weight | Max marks | Excellent | Average | Poor |
|---|---|---|---|---|---|
| Concept correctness | 20% | 10 | Correctly identifies master-slave triggering mechanism in (a), quarter-wave transformer impedance transformation principle in (b), induction motor power flow hierarchy in (c), AM modulation theory including sideband generation in (d), and proper Y-to-ABCD/h-parameter conversion relationships in (e) | Shows partial understanding with minor conceptual errors such as confusing level vs edge triggering, misapplying transmission line formulas, or incorrect power flow sequence in motor; parameter conversions mostly correct but with sign errors | Fundamental misconceptions like treating master-slave as simultaneous triggering, using wrong impedance formula for transmission lines, confusing stator and rotor losses, or attempting direct parameter assignment without conversion |
| Numerical accuracy | 20% | 10 | All calculations precise: (b) Z_in = 60 Ω, (c) rotor Cu loss = 1.54 kW, shaft power = 36.16 kW and at reduced speed = 14.56 kW, (d) m = 0.2, sideband amplitude = 5 V, BW = 1 kHz, power = 2.55 W, (e) exact ABCD and h-parameter values with correct units | Correct approach with minor arithmetic errors (e.g., slip calculation error in (c), power formula mistakes, or parameter conversion arithmetic slips); final answers slightly off but method evident | Major calculation errors, missing units, order-of-magnitude mistakes, or omitted numerical answers despite correct formulas; significant errors in handling per-unit values or decibel conversions |
| Diagram quality | 20% | 10 | Clear timing diagram for (a) showing clock, S, R, master output (Q_m), and slave output (Q) with proper delays and edge alignment; neat frequency spectrum for (d) with labeled axes, carrier and sideband amplitudes, and frequency markers; labeled circuit diagrams if drawn | Diagrams present but with minor timing misalignments, missing labels, or unclear amplitude representations; frequency spectrum lacks proper scaling or omits some frequency labels | Missing diagrams, extremely rough sketches without axes labels, incorrect timing relationships showing simultaneous rather than delayed outputs, or completely wrong spectrum representation |
| Step-by-step derivation | 20% | 10 | Each sub-part shows complete derivation chain: (a) explains why slave follows master on falling edge; (b) states Z_in = Z₀²/Z_L before substitution; (c) clearly shows P_g → sP_g → (1-s)P_g → P_shaft flow; (d) derives m, then sidebands, then power stepwise; (e) shows matrix inversion or substitution method for parameter conversion | Some steps implied or skipped, particularly in parameter conversions or power flow equations; final answers present but intermediate steps condensed or unclear | No derivations shown, only final answers stated; or incorrect formulas applied without explanation; jumps from given data to final answer without logical connection |
| Practical interpretation | 20% | 10 | Interprets results practically: explains why master-slave eliminates race conditions in (a); notes that matched line shows frequency-independent Z_in in (b); discusses efficiency implications of high slip operation in (c); relates low m=0.2 to power efficiency trade-offs in AM broadcasting in (d); comments on network reciprocity or symmetry in (e) | Brief mention of practical significance without elaboration; standard concluding statements without specific connection to calculated values | No interpretation or discussion of results; purely mathematical treatment without physical or engineering context; misses opportunity to discuss real-world implications of key findings |
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