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Electrical Knowledge Flashcards

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

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  1. A 480V, 3-phase motor draws 18A under full load. If the power factor is 0.82 lagging, what is the motor's approximate true power consumption?

    Answer: 12.2 kW

    True power (kW) = √3 × V × I × PF = 1.732 × 480 × 18 × 0.82 ≈ 12,224W ≈ 12.2 kW. The power factor of 0.82 represents the ratio of true (real) power to apparent power, so multiplying the apparent power by 0.82 gives the actual working power consumed.

  2. When performing insulation resistance testing with a megohmmeter on a motor winding, a technician gets a reading of 0.5 MΩ after one minute. According to standard practice, this motor should be:

    Answer: Taken out of service — the reading indicates dangerously degraded insulation

    The industry rule of thumb (IEEE 43) requires a minimum of 1 MΩ plus 1 MΩ per kV of operating voltage for the motor to be considered safe to energize. A reading of 0.5 MΩ falls below even the baseline 1 MΩ threshold, indicating moisture intrusion, contamination, or insulation breakdown — the motor must be taken out of service.

  3. In a series RLC circuit operating below its resonant frequency, the circuit's net reactive behavior is:

    Answer: Capacitive, because XC > XL

    At frequencies below resonance, the capacitive reactance (XC = 1/2πfC) is greater than the inductive reactance (XL = 2πfL). Because XC dominates, the net impedance is capacitive, current leads voltage, and the circuit behaves like a capacitor. At resonance XL = XC; above resonance the circuit becomes inductive.

  4. A technician measures 120V between L1 and ground, 120V between L2 and ground, but only 120V between L1 and L2 on a system that should be 240V single-phase. The most likely cause is:

    Answer: Both hot conductors are in phase — there is no phase shift between them

    In a normal 240V single-phase center-tap system, L1 and L2 are 180° out of phase, so the voltage between them is 120 + 120 = 240V. Reading only 120V between L1 and L2 means the two hots are in phase (0° apart), not 180° apart — a classic sign of incorrect transformer wiring or a single-phase supply fed from two conductors of the same phase. An open neutral would cause imbalanced loads, not this phase symptom.

  5. According to the NEC, the maximum allowable voltage drop for a combination of feeder and branch circuit conductors serving a sensitive load is:

    Answer: 5% total — with no prescribed split between feeder and branch

    NEC 210.19(A) Informational Note No. 4 and 215.2(A)(1) Informational Note No. 2 recommend (not mandate) that the combined voltage drop for feeders and branch circuits not exceed 5% total for sensitive electronic loads. The NEC itself does not prescribe a mandatory split between feeder and branch — the 3%/2% split is a common design recommendation from FPN, but the hard guidance is a 5% combined maximum.

  6. A 4–20 mA current loop transmitter is wired to a PLC analog input card with a 250 Ω precision resistor across its terminals. At a process value representing 75% of span, what voltage should the technician measure across the resistor?

    Answer: 4.00V

    At 75% of span the current is: 4 mA + 0.75 × (20 – 4) mA = 4 + 12 = 16 mA. Voltage across the 250 Ω resistor = 0.016 A × 250 Ω = 4.00V. This is a common troubleshooting calculation on 4–20 mA instrumentation loops; confusing the span (16 mA) with the total (20 mA) or forgetting the 4 mA live-zero offset are the two most frequent errors.