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UPSC IES/ESE (Engineering Services) Electrical Machines, Power Systems and Power Electronics Flashcards

60 question-and-answer cards covering Electrical Machines, Power Systems and Power Electronics as it is examined in UPSC IES/ESE (Engineering Services). 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

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24 sample cards from the Electrical Machines, Power Systems and Power Electronics deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

  1. Define voltage regulation of a transformer.

    $\%\,\text{Regulation} = \dfrac{V_{no\text{-}load} - V_{full\text{-}load}}{V_{full\text{-}load}} \times 100$. Approximately $\%R \approx V_R\cos\phi \pm V_X\sin\phi$ (+ for lagging, − for leading p.f.).

  2. State the advantage of an auto-transformer over a two-winding transformer.

    It uses a single tapped winding (part conducted, part transformed), so it requires less copper, has lower losses, higher efficiency, better regulation and is cheaper. Saving of copper $\propto (1-k)$ where $k=N_2/N_1$. Drawback: no electrical isolation between primary and secondary.

  3. State the conditions for satisfactory parallel operation of single-phase transformers.

    (1) Same voltage ratings / turns ratio. (2) Same polarity. (3) Same per-unit (percentage) impedance. (4) Same X/R ratio. For three-phase: also same phase sequence and same phase displacement (vector group).

  4. Which two tests determine transformer losses, and which loss does each give?

    Open-circuit (no-load) test gives core/iron losses (and shunt branch parameters); short-circuit test gives full-load copper losses (and series impedance).

  5. Explain why the open-circuit test is done on the LV side and short-circuit test on the HV side.

    OC test on LV side: rated low voltage is easier to apply and the meters carry small no-load current. SC test on HV side: rated current flows at a low applied voltage (a few percent of rated), convenient and safe for measurement.

  6. State the working principle of a three-phase induction motor.

    Three-phase currents produce a rotating magnetic field (rotating at synchronous speed). This field induces (by Faraday's law) EMF and currents in the rotor; the interaction of rotor currents with the field produces torque (Lenz's law) that drags the rotor along. It is essentially a rotating transformer.

  7. Define slip of an induction motor and give synchronous speed.

    $N_s = \dfrac{120 f}{P}$ (rpm). Slip $s = \dfrac{N_s - N}{N_s}$, often as a percentage. Rotor frequency $f_r = s f$.

  8. Give the condition for maximum torque in a three-phase induction motor.

    Maximum (pull-out) torque occurs when rotor resistance equals standstill rotor reactance referred per phase: $\dfrac{R_2}{s} = X_2$, i.e. slip at maximum torque $s_m = \dfrac{R_2}{X_2}$. Maximum torque is independent of $R_2$.

  9. List the common starting methods for three-phase induction motors.

    Squirrel-cage: Direct-On-Line (DOL), Star–Delta starter, Auto-transformer starter, Soft starter / rotor-resistance for slip-ring. Slip-ring motors: add external rotor resistance to boost starting torque and limit current.

  10. Explain speed control of induction motors by varying supply frequency (V/f control).

    Since $N_s = \dfrac{120f}{P}$, varying f changes speed. To keep flux constant and avoid saturation, voltage is varied proportionally so that $\dfrac{V}{f}$ is held constant — the basis of VFD/V-f control.

  11. Name methods of braking an induction motor.

    (1) Regenerative braking (rotor driven above synchronous speed, machine acts as generator). (2) Plugging (reverse phase sequence). (3) Dynamic (rheostatic) braking — DC injected into stator.

  12. Write the voltage regulation expression for an alternator (synchronous generator) and define it.

    $\%\,\text{Reg} = \dfrac{E_0 - V}{V}\times100$, where $E_0$ is no-load (excitation) EMF and V is terminal voltage at full load. $E_0 = \sqrt{(V\cos\phi + I R_a)^{2} + (V\sin\phi \pm I X_s)^{2}}$ (+ lagging, − leading).

  13. List methods of finding voltage regulation of a synchronous generator.

    (1) Synchronous impedance / EMF method (gives pessimistic value). (2) Ampere-turn / MMF method (gives optimistic value). (3) Zero power factor / Potier method. (4) ASA method.

  14. How is power factor of a synchronous motor controlled, and what is a synchronous condenser?

    By varying field (DC) excitation: over-excitation makes it draw leading current (capacitive), under-excitation makes it draw lagging current. The V-curves show this. A synchronous condenser is an over-excited synchronous motor running on no-load used to supply leading reactive power and improve system power factor.

  15. State the synchronization conditions for connecting an alternator to the grid.

    (1) Same voltage magnitude. (2) Same frequency. (3) Same phase sequence. (4) Same phase (in phase) at the instant of paralleling.

  16. Define load factor, demand factor and diversity factor in power systems.

    Load factor $= \dfrac{\text{average load}}{\text{maximum demand}}$. Demand factor $= \dfrac{\text{maximum demand}}{\text{connected load}}$. Diversity factor $= \dfrac{\text{sum of individual max demands}}{\text{coincident max demand of group}}\ (\geq 1)$.

  17. What is the most economical conductor cross-section for a transmission line (Kelvin's law)?

    Kelvin's law: the most economical conductor size is that for which the annual cost of energy loss (variable, $\propto$ area) equals the annual interest plus depreciation on the capital cost of the conductor (fixed, $\propto$ area).

  18. Define inductance and capacitance per phase of a transmission line (single conductor).

    Inductance per phase $L = 2\times10^{-7}\ln\dfrac{D}{r'}\ \text{H/m}$ (with $r' = r e^{-1/4}$ = GMR). Capacitance per phase $C = \dfrac{2\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}$, where D = spacing between conductors.

  19. Compare short, medium and long transmission line models.

    Short ($<80$ km, $<$ 11–66 kV): only series R and L; capacitance neglected. Medium (80–250 km): capacitance included as lumped (nominal-T or nominal-$\pi$). Long ($>250$ km): distributed parameters, solved with hyperbolic functions / ABCD constants.

  20. For a long transmission line, give the characteristic impedance and propagation constant.

    Characteristic (surge) impedance $Z_c = \sqrt{\dfrac{z}{y}} = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$; propagation constant $\gamma = \sqrt{zy} = \alpha + j\beta$. For a lossless line $Z_c = \sqrt{L/C}$.

  21. Define the Ferranti effect.

    On a long or medium transmission line under no-load or light-load conditions, the receiving-end voltage rises above the sending-end voltage due to the line's charging (capacitive) current flowing through the line inductance.

  22. What is corona, and give Peek's formula for critical disruptive voltage.

    Corona is the luminous discharge (ionization of air) around a conductor when the surface electric field exceeds the breakdown strength of air, causing power loss, hissing and ozone. Critical disruptive voltage $V_d = m_0\,\delta\,g_0\,r\ln\dfrac{D}{r}$, where $g_0\approx30\ \text{kV/cm}$, $\delta$ = air density factor, $m_0$ = conductor surface factor.

  23. List factors affecting corona loss and ways to reduce it.

    Affected by: supply voltage, frequency, conductor diameter, spacing, air density, and surface condition. Reduced by using larger-diameter conductors, bundled conductors, hollow/ACSR conductors, and increasing spacing. Peek's corona loss $\propto (f+25)(V-V_d)^{2}$.

  24. Define string efficiency of a suspension insulator and how to improve it.

    $\text{String efficiency} = \dfrac{\text{voltage across the whole string}}{n \times \text{voltage across the disc nearest the conductor}}$. Improved by using a guard/grading ring, longer cross-arms, and capacitance grading. Ideal string efficiency is $100\%$ (equal voltage on all discs).

What this deck covers

The Electrical Machines, Power Systems and Power Electronics deck follows the UPSC IES/ESE (Engineering Services) Electrical Machines, Power Systems and Power Electronics syllabus — 6 chapters and 24 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 10.0 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 227 characters, which is long enough to carry the reasoning and short enough to say out loud.

A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.

Electrical Machines, Power Systems and Power Electronics flashcards FAQ

How many Electrical Machines, Power Systems and Power Electronics flashcards are in this UPSC IES/ESE (Engineering Services) deck?

60 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.

Are these UPSC IES/ESE (Engineering Services) flashcards free?

Yes. The preview here is free to read with no signup, and the full 60-card deck is free inside the Examius app.

What do the Electrical Machines, Power Systems and Power Electronics cards cover?

They follow the UPSC IES/ESE (Engineering Services) Electrical Machines, Power Systems and Power Electronics syllabus — 6 chapters and 24 topics — so the questions track what is actually examinable.

How should I use these flashcards?

Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.