Paper I — Q5
(a) A step down dc chopper is feeding a load of R = 10 Ω and L = 20 mH. The dc supply voltage is 100 V. The chopper is switching…
A step down dc chopper is feeding a load of R = 10 Ω and L = 20 mH. The dc supply voltage is 100 V. The chopper is switching at a frequency of 2 kHz with a duty cycle of 50%. Determine the load current and the peak-to-peak ripple current as an absolute value and as percentage of dc value. 10 marks
In a certain material with σ = 0, ε = ε₀ εᵣ and μ = μ₀ μᵣ, the magnetic field intensity component is given by H = 10 sin (10⁸ t – 2x) aᵤ A/m. Find the following: (i) Displacement current density (ii) Electric field intensity 10 marks
A Scott connected transformer shown in Figure 5(c) is supplied from 11 kV, 3-phase, 50 Hz mains. Secondaries are series connected and supply 1100 A at a voltage of 100√2 V to a resistive load. The phase sequence of the 3-phase supply is ABC. (i) Calculate the turns ratio of the teaser transformer. (ii) Calculate the line current I_B and its phase angle with respect to the voltage of phase A to neutral on the 3-phase side. 10 marks
A transmitter with a 10 kW carrier transmits 11·2 kW when modulated with a single sine wave. Calculate the modulation index. If the carrier is simultaneously modulated with two other sine waves also at 50% modulation, calculate the total power transmitted. 10 marks
For the circuit shown in Figure 5(e), v_C(0+) = 2 V and i(0+) = 2/3 A. Calculate the value of v_C(t) for t > 0. 10 marks
हिंदी में प्रश्न पढ़ें
एक अपचायी dc संकटक (चॉपर), एक R = 10 Ω तथा L = 20 mH वाले भार को पोषित करता है । dc प्रदाय वोल्टता 100 V है । संकटक 50% कर्म चक्र के साथ 2 kHz की आवृत्ति पर स्विचिंग कर रहा है । भार धारा तथा शिखर-से-शिखर ऊर्मिका धारा का मान, विशुद्ध मान व dc मान के प्रतिशत के रूप में ज्ञात कीजिए । (10 अंक)
एक किसी σ = 0, ε = ε₀ εᵣ तथा μ = μ₀ μᵣ वाले पदार्थ में चुंबकीय क्षेत्र की तीव्रता का घटक H = 10 sin (10⁸ t – 2x) aᵤ A/m द्वारा प्रदर्शित है । निम्नलिखित का मान ज्ञात कीजिए : (i) विस्थापन धारा घनत्व (ii) विद्युत क्षेत्र तीव्रता (10 अंक)
चित्र 5(c) में प्रदर्शित एक स्कॉट संयोजित परिणामित्र, 11 kV, त्रि-कला, 50 Hz मैंस से पोषित है । द्वितीयक श्रेणी क्रम में संयोजित हैं और एक प्रतिरोधीय भार को 100√2 V पर 1100 A प्रदान करती है । त्रि-कला प्रदाय (सप्लाई) का कला क्रम ABC है । (i) टीज़र परिणामित्र का फेरा अनुपात परिकलित कीजिए । (ii) त्रि-कला दिशा में तत्क्ष से कला A की बोल्टता के सापेक्ष लाइन धारा I_B तथा इसके कला कोण की गणना कीजिए । (10 अंक)
एक प्रेषित्र 10 kW वाहक के साथ 11·2 kW प्रेषित करता है जबकि एकल ज्या तरंग द्वारा मॉडुलित होता है । मॉडुलन सूचकांक की गणना कीजिए । यदि वाहक साथ ही साथ दो अन्य ज्या तरंगों द्वारा भी 50% मॉडुलन पर मॉडुलित होता है, तो संपूर्ण प्रेषित शक्ति की गणना कीजिए । (10 अंक)
चित्र 5(e) में प्रदर्शित परिपथ के लिए v_C(0+) = 2 V तथा i(0+) = 2/3 A हैं । t > 0 के लिए v_C(t) के मान की गणना कीजिए । (10 अंक)
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.
(c) Figure 5(c) shows a Scott connected transformer circuit. On the left, a 3-phase supply labeled '11 kV, 3-phase supply' has three terminals: A, B, and C. Terminal A connects to the top of a vertical inductor labeled 'Teaser' with a turns ratio of (sqrt(3)/2)N1. Terminal B connects to a horizontal inductor labeled N1/2, which connects to a central node M. From node M, another horizontal inductor labeled N1/2 connects to terminal C. This arrangement is labeled 'Main'. On the right side, there is a vertical inductor labeled N2 connected in series with the top of the Teaser inductor. The bottom of the Teaser inductor connects to node M. The bottom of the Main transformer section (from the N1/2 inductor connected to C) connects to the bottom of a second vertical inductor labeled N2. The top of the right-side N2 inductor and the bottom of the left-side N2 inductor form the secondary output terminals. These terminals are connected to a 'Resistive load'. The secondary voltage is indicated as 100/sqrt(2) V (with a vertical double-headed arrow) and the current is indicated as 1100 A (with a horizontal arrow pointing right).
(e) Circuit diagram labeled Figure 5(e). The circuit consists of a capacitor C1 (0.5 F) on the left, a dependent current source in the middle, a resistor (3 Ohm) in series with a capacitor C2 (1/3 F) on the right. The components are connected in a parallel configuration between a top and bottom rail. The voltage across C1 is labeled vC with the positive terminal at the top. The dependent current source is a diamond shape with an arrow pointing upwards, labeled with the value i/2. The current flowing through the right branch (through the 3 Ohm resistor and C2) is labeled i, with an arrow pointing upwards. The bottom rail is a common ground line.
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) Chopper load current and ripple Assume continuous conduction with a freewheeling diode. V₀ = DV_s = 0.5 × 100 = 50 V, so I₀ = 50/10 = 5 A. T = 0.5 ms, t_on = t_off = 0.25 ms, τ = L/R = 2 ms. I_max = (V_s/R)(1 − e^(−t_on/τ))/(1 − e^(−T/τ)) = 10 × 0.1175/0.2212 = 5.3125 A I_min = (V_s/R)(e^(t_on/τ) − 1)/(e^(T/τ) − 1) = 10 × 0.1331/0.2840 = 4.6875 A Peak-to-peak ripple = 0.625 A, which is 12.5% of the dc current. The approximation V_sD(1 − D)/(fL) = 100 × 0.25/(2000 × 0.02) = 0.625 A agrees; the ripple is largest at D = 0.5.
(b) Displacement current density and electric field H = 10 sin(10⁸t − 2x) a_z A/m, σ = 0, ω = 10⁸ rad/s, β = 2 rad/m (wave along +x). (i) With σ = 0, ∇ × H = J_d. For H = H_z(x, t) a_z, ∇ × H = −(∂H_z/∂x) a_y = 20 cos(10⁸t − 2x) a_y, so J_d = 20 cos(10⁸t − 2x) a_y A/m². (ii) J_d = ε ∂E/∂t, so E = (1/ε)∫J_d dt = [20/(10⁸ε₀εᵣ)] sin(10⁸t − 2x) a_y. With ε₀ = 10⁻⁹/(36π) F/m, E = (7200π/εᵣ) sin(10⁸t − 2x) a_y ≈ (22.6 × 10³/εᵣ) sin(10⁸t − 2x) a_y V/m. Since β = ω√(μ₀μᵣε₀εᵣ) gives μᵣεᵣ = 36, the same field is E = 200πμᵣ sin(10⁸t − 2x) a_y V/m; E × H points along +x, as it should.
(c) Scott-connected transformer Assumption: the figure shows no polarity dots on the main transformer, so its secondary is taken in phase with V_BC; the teaser secondary is in phase with V_AM, which is in phase with V_AN. The two secondary voltages are then in quadrature. (i) The series sum is √2 V₂ = 100√2 V, so each secondary gives V₂ = 100 V. The teaser primary sees V_AM = (√3/2) × 11 000 = 9526 V. Teaser turns ratio = 9526 : 100 = 95.26 : 1 (main: 110 : 1). (ii) Load voltage = 100∠0° + 100∠−90° = 100√2∠−45° V, so I₂ = 1100∠−45° A (resistive load). Teaser primary: I_A = 1100∠−45°/95.26 = 11.55∠−45° A, shared equally by the two halves of the main winding (−5.77∠−45° A each into B and C). Main primary: 1100∠−45°/110 = 10∠−45° A, from B to C. I_B = 10∠−45° − 5.77∠−45° = 4.23 A at −45°, i.e. lagging V_AN by 45°. (I_C = 15.77∠135° A; the sum I_A + I_B + I_C is zero. With the main secondary reversed, I_B = 15.77 A at −135°.)
(d) AM power Single tone: P_t = P_c(1 + m²/2), so 11.2 = 10(1 + m²/2), m² = 0.24 and m = 0.49. Adding two tones at 50%, the sideband powers add: m_t² = 0.24 + 0.25 + 0.25 = 0.74 (m_t ≈ 0.86). P_t = 10(1 + 0.74/2) = 13.7 kW. The carrier stays at 10 kW; the sideband power rises from 1.2 kW to 3.7 kW.
(e) Transient of v_C Top node: the dependent source (i/2) and the right branch (i) both feed C₁ = 0.5 F, so 0.5 dv_C/dt = 1.5i, or dv_C/dt = 3i. Right branch: v₂ = v_C + 3i, and i = −C₂ dv₂/dt = −(1/3)(dv_C/dt + 3 di/dt) = −i − di/dt, so di/dt + 2i = 0. i = (2/3)e^(−2t) A, dv_C/dt = 2e^(−2t) V/s and v_C(0⁺) = 2 V, so v_C(t) = 3 − e^(−2t) V, t > 0.
How this answer will be evaluated
What earns full marks: All parts solved with correct methods, clear derivations, and proper units.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Average load current and peak-to-peak ripple current (absolute and as a percentage of the dc value) of a step-down chopper feeding an R–L load. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Find V₀ = DV_s = 50 V and I₀ = 5 A.
- Find I_max = 5.3125 A and I_min = 4.6875 A using τ = L/R = 2 ms.
- State the ripple as 0.625 A, i.e. 12.5% of the dc current.
Loses marks
- Wrong average current or ignoring the freewheeling diode.
- Confusing peak-to-peak with peak ripple.
Earns more
- ON and OFF equations with the freewheeling diode stated.
- Linear approximation V_sD(1 − D)/(fL) checked.
Extra mark
- Load-current waveform sketched with the ripple.
- (b) Displacement current density and electric field intensity for a given H in a lossless medium. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use ∇ × H = J_d (σ = 0) to get J_d = 20 cos(10⁸t − 2x) a_y A/m².
- Integrate J_d = ε ∂E/∂t to get E = [20/(10⁸ε₀εᵣ)] sin(10⁸t − 2x) a_y ≈ (22.6 × 10³/εᵣ) sin(10⁸t − 2x) a_y V/m.
- Check direction (E × H along +x) and units.
Loses marks
- Sign or direction error in the curl.
- Missing units or ε₀ left out in E.
Earns more
- Curl taken correctly for H along z varying with x.
- Use of β = 2 to relate μᵣεᵣ = 36.
Extra mark
- Equivalent form E = 200πμᵣ sin(10⁸t − 2x) a_y V/m.
- (c) Teaser turns ratio of the Scott-connected transformer and the line current I_B with its phase angle. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Find V₂ = 100 V per secondary from the quadrature sum 100√2 V and the teaser ratio 9526 : 100 = 95.26 : 1.
- Find the load current 1100∠−45° A and reflect it to the teaser (11.55 A) and the main (10 A) primaries.
- Combine at M to get I_B = 4.23 A at −45° with respect to V_AN, with the polarity assumption stated.
Loses marks
- Using 11 kV rather than (√3/2) × 11 kV across the teaser primary.
- Confusing line and phase currents or omitting the phase angle.
Earns more
- Teaser primary voltage (√3/2) × 11 kV used.
- Check that I_A + I_B + I_C = 0.
Extra mark
- Phasor diagram of voltages and currents.
- (d) Modulation index from the power data and the total power with two further tones at 50% modulation. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use P_t = P_c(1 + m²/2) to get m = 0.49.
- Add the squared indices of the tones: m_t² = 0.24 + 0.25 + 0.25 = 0.74.
- Compute P_t = 13.7 kW.
Loses marks
- Adding the indices directly (0.49 + 0.5 + 0.5) instead of in quadrature of power.
- Confusing carrier and sideband power.
Earns more
- Carrier power constant, sideband power rising from 1.2 kW to 3.7 kW.
- Correct units.
Extra mark
- Check that m_t < 1 (no overmodulation).
- (e) v_C(t) for t > 0 in a circuit with two capacitors and a dependent current source. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Write KCL at the top node: 0.5 dv_C/dt = i/2 + i, giving dv_C/dt = 3i.
- Write the right-branch equations to get di/dt + 2i = 0 and i = (2/3)e^(−2t) A.
- Integrate with v_C(0⁺) = 2 V to get v_C(t) = 3 − e^(−2t) V.
Loses marks
- Missing the dependent-source current in KCL.
- Sign error in i = −C₂ dv₂/dt.
Earns more
- Correct current directions and capacitor sign conventions.
- Initial conditions applied.
Extra mark
- Final value 3 V explained as the charge pumped by the dependent source.
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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