Electrical Knowledge Flashcards

A 3-phase, 480V motor draws 18A per phase at a power factor of 0.78 lagging. What is the true power consumed by the motor?

Answer: 11,655 W

True power (watts) = √3 × V × I × PF = 1.732 × 480 × 18 × 0.78 ≈ 11,655 W. The power factor of 0.78 lagging means only 78% of apparent power is doing real work. Forgetting to multiply by √3 for 3-phase or omitting the PF are the most common errors.

When using a megohmmeter (megger) to test insulation resistance on a motor winding, you obtain a reading of 0.8 MΩ after 1 minute. According to IEEE 43, this reading indicates:

Answer: Insulation is questionable and the motor should be investigated before energizing

IEEE 43 recommends a minimum insulation resistance of 1 MΩ for motors rated below 1 kV. A reading of 0.8 MΩ falls below this threshold, indicating questionable insulation that warrants further investigation (cleaning, drying, or repair) before returning the motor to service. It does not automatically mean the winding must be rewound, but energizing it risks failure.

Two capacitors, 40 µF and 60 µF, are connected in series across a 120V AC source. What voltage appears across the 40 µF capacitor?

Answer: 72 V

In a series capacitive circuit, voltage divides inversely proportional to capacitance. The series capacitance is (40×60)/(40+60) = 24 µF. Charge Q = C_total × V = 24 µF × 120V = 2880 µC. V across 40 µF = Q / C = 2880 / 40 = 72V. The smaller capacitor drops the larger voltage — the opposite of series resistors.

A technician measures 277V between a phase conductor and ground in a 480/277V wye system. When he measures between the same phase conductor and the neutral bar in the panel, he reads 0V. The most likely cause is:

Answer: The neutral conductor is open (broken) between the panel and the service entrance

In a properly functioning 480/277V wye system, phase-to-neutral voltage should be 277V. If phase-to-ground reads 277V but phase-to-neutral reads 0V, the neutral bar and ground are at the same potential at the panel — meaning the neutral conductor is open upstream. The panel's ground and neutral are both referencing earth through the grounding electrode, masking the fault on the voltage-to-ground measurement but revealing 0V between phase and the floating/open neutral bar.

According to NEC Article 430, the maximum allowable dual-element time-delay fuse size for overload protection of a 10 HP, 230V, single-phase motor with a nameplate full-load current of 50A and a service factor of 1.15 is:

Answer: 75 A

NEC 430.52 sets the maximum branch-circuit short-circuit and ground-fault protection. For overload protection using a dual-element time-delay fuse on a motor with a service factor ≥ 1.15, NEC 430.32(A)(1) allows up to 125% of FLA = 50A × 1.25 = 62.5A. However, if this size is insufficient to allow the motor to start, it may be increased up to 140% per NEC 430.32(C) = 50 × 1.40 = 70A — and the next standard fuse size up is 75A. The question asks for the maximum allowable, which follows the 140% path rounded to the next standard size: 75A.

In a DC motor, if the field winding circuit opens while the motor is running under full load, what will most likely occur?

Answer: The motor speed will increase dramatically, potentially to a dangerous overspeed condition

In a DC shunt or series motor, the field flux provides the counter-EMF that limits armature current and governs speed. If the field opens, magnetic flux drops toward zero. To maintain torque against the load, the armature current spikes — but more critically, with near-zero flux, back-EMF collapses and the speed equation (N ∝ V / Φ) causes speed to rise dramatically. This 'runaway' condition can mechanically destroy the motor. This is why DC motors (especially series motors) should never be run unloaded or with an open field.