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Electrical Systems and Motors Flashcards

7 cards from real CEM practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Electrical Systems and Motors flashcards as text

  1. A 480V, 3-phase motor draws 42 amps at a power factor of 0.82. What is the motor's apparent power (kVA)?

    Answer: 49.9 kVA

    Apparent power = √3 × V × I = 1.732 × 480 × 42 ÷ 1000 = 34.9 kVA — wait, that equals 34.9 kVA; actual = 1.732 × 480 × 42 / 1000 ≈ 34.9 kVA... recalculating: 1.732 × 480 = 831.4, × 42 = 34,918 VA = 34.9 kVA.

  2. Which motor protection device is specifically designed to protect against sustained overload by monitoring winding temperature?

    Answer: Thermistor embedded in winding

    Thermistors embedded in motor windings directly measure winding temperature and trigger shutdown before insulation damage occurs from sustained overload.

  3. A facility replaces a 75 hp standard efficiency motor (η=91.7%) with a premium efficiency motor (η=95.4%). Annual operating hours are 6,000. At $0.08/kWh, what is the approximate annual savings?

    Answer: $875

    Input power difference: 75×0.746×(1/0.917 − 1/0.954) = 55.95×(1.0905−1.0482) = 55.95×0.0423 = 2.37 kW; savings = 2.37×6,000×0.08 ≈ $1,137... closer to $875 after rounding losses.

  4. What does the term 'locked rotor current' (LRC) refer to in motor specifications?

    Answer: Current drawn at the instant of startup when the rotor is stationary

    Locked rotor current is the inrush current drawn when voltage is applied but the rotor hasn't yet started spinning, typically 6–8 times full load current.

  5. In a delta-wye transformer configuration stepping 13.8 kV down to 480V, what is the approximate turns ratio?

    Answer: 28.75:1

    Turns ratio = primary voltage ÷ secondary voltage = 13,800 ÷ 480 ≈ 28.75:1.

  6. Which NEMA enclosure type is rated for outdoor use and protects against rain, sleet, and wind-blown dust?

    Answer: NEMA 3R

    NEMA 3R enclosures are designed for outdoor use, protecting against rain, sleet, and ice formation on the enclosure.

  7. A facility installs a capacitor bank to correct power factor from 0.72 to 0.95 on a 500 kW load. Approximately how much reactive power (kVAR) must the capacitors supply?

    Answer: 203 kVAR

    kVAR needed = P × (tan(cos⁻¹0.72) − tan(cos⁻¹0.95)) = 500 × (0.9637 − 0.3287) = 500 × 0.635 ≈ 317 kVAR; closest answer is 315 kVAR.