Master Electrician (NEC) — All Questions

A load draws 80 kW of real power and 100 kVA of apparent power. What is the power factor?

• a.1.0
• b.0.6
• c.0.8✓
• d.0.9

Power factor = real power / apparent power = 80 kW / 100 kVA = 0.8.

For a load of 80 kW real and 100 kVA apparent, what is the reactive power?

• a.80 kVAR
• b.60 kVAR✓
• c.40 kVAR
• d.100 kVAR

Reactive power = sqrt(kVA^2 - kW^2) = sqrt(100^2 - 80^2) = sqrt(3,600) = 60 kVAR.

What is the real power of a three-phase load at 480 V, 50 A, and 0.85 power factor?

• a.35,326 W✓
• b.24,960 W
• c.30,000 W
• d.41,568 W

P = 1.732 x V x I x PF = 1.732 x 480 x 50 x 0.85 = 35,326 W.

In a balanced wye (star) three-phase system, the line current is:

• a.0.577 times the phase current
• b.Equal to the phase current✓
• c.1.732 times the phase current
• d.2 times the phase current

In a wye connection the same conductor carries phase and line current, so line current equals phase current.

In a balanced delta three-phase system, the line current equals:

• a.3 times the phase current
• b.0.577 times the phase current
• c.The phase current
• d.1.732 times the phase current✓

In a delta connection the line current is the square root of 3 (1.732) times the phase (winding) current.

In a 208Y/120-V wye system, what is the phase (line-to-neutral) voltage?

• a.208 V
• b.120 V✓
• c.277 V
• d.240 V

Line-to-neutral voltage = line-to-line / 1.732 = 208 / 1.732 = 120 V.

In a 480Y/277-V system, what is the line-to-neutral voltage?

• a.240 V
• b.120 V
• c.277 V✓
• d.208 V

Line-to-neutral = 480 / 1.732 = 277 V.

An 80-kW load at 0.80 power factor is to be corrected to 0.95. How many kVAR of capacitance are required (tan 36.87 = 0.75, tan 18.19 = 0.329)?

• a.48 kVAR
• b.60 kVAR
• c.33.7 kVAR✓
• d.20 kVAR

Qc = kW x (tan(theta1) - tan(theta2)) = 80 x (0.750 - 0.329) = 80 x 0.421 = 33.7 kVAR.

In a three-phase, four-wire wye system serving nonlinear loads, which harmonic order adds arithmetically in the neutral?

• a.7th
• b.2nd
• c.5th
• d.3rd✓

Triplen harmonics, principally the 3rd, are in phase across all three phases and add in the neutral rather than canceling.

On a feeder with significant harmonic (nonlinear) load, the neutral conductor is treated as:

• a.Not a current-carrying conductor
• b.A current-carrying conductor✓
• c.An ignored conductor
• d.A half current-carrying conductor

Where the major portion of the load is nonlinear, the neutral carries harmonic current and must be counted as a current-carrying conductor for adjustment.2023 NEC 310.15(E)

A three-phase panel has phase loads of A = 40 A, B = 30 A, C = 20 A. What is the ideal balanced current per phase?

• a.20 A
• b.40 A
• c.25 A
• d.30 A✓

Total = 40 + 30 + 20 = 90 A; balanced across three phases = 90 / 3 = 30 A per phase.

A single-line (one-line) diagram represents an electrical system by:

• a.Only the branch-circuit wiring
• b.The physical layout of raceways
• c.One line for each circuit or set of conductors✓
• d.Each individual conductor drawn separately

A one-line diagram uses a single line and standard symbols to represent circuits and equipment, simplifying analysis of the power system.

A transformer steps 480 V down to 120 V. What is its turns (voltage) ratio?

• a.4:1✓
• b.8:1
• c.2:1
• d.1:4

Turns ratio = primary voltage / secondary voltage = 480 / 120 = 4:1.

A transformer has a 4:1 turns ratio and a 10-A primary current. What is the secondary current (ideal)?

• a.2.5 A
• b.40 A✓
• c.10 A
• d.20 A

Current is inversely proportional to the turns ratio: Is = Ip x (Np/Ns) = 10 x 4 = 40 A.

An AC circuit at 240 V draws 12 A. What is the circuit impedance?

• a.0.05 ohms
• b.2 ohms
• c.20 ohms✓
• d.288 ohms

By Ohm's law for AC, Z = V / I = 240 / 12 = 20 ohms.

A continuous load draws 40 A. At what minimum ampacity must the branch-circuit conductors and overcurrent device be rated?

• a.50 A✓
• b.32 A
• c.40 A
• d.45 A

Continuous loads require 125% sizing: 40 A x 1.25 = 50 A for the conductors and overcurrent device.2023 NEC 210.20(A)

The demand factor of a system is defined as:

• a.Connected load divided by maximum demand
• b.A value always greater than 1
• c.Maximum demand divided by total connected load✓
• d.Always equal to 1

Demand factor = maximum demand / total connected load; it is 1 or less and reflects load diversity.

What are the commonly recommended maximum voltage-drop limits for a branch circuit and for the combined feeder plus branch circuit?

• a.2% and 4%
• b.3% and 5%✓
• c.5% and 8%
• d.1% and 3%

The informational recommendation is 3% maximum on a branch circuit and 5% maximum for feeder plus branch circuit combined.

What three-phase transformer kVA rating corresponds to a full-load current of 100 A at 480 V?

• a.48 kVA
• b.144 kVA
• c.83 kVA✓
• d.100 kVA

kVA = (1.732 x V x I) / 1000 = (1.732 x 480 x 100) / 1000 = 83.1 kVA. Trap: 48 kVA omits the 1.732 three-phase factor.

A single-phase load draws 30 A at 240 V with a 0.90 power factor. What is the real power?

• a.7,200 W
• b.6,480 W✓
• c.5,760 W
• d.6,000 W

P = V x I x PF = 240 x 30 x 0.90 = 6,480 W. Trap: 7,200 W is the apparent power (VA) before power factor is applied.

A resistive heating element carries 10 A and has a resistance of 2 ohms. What power does it dissipate?

• a.20 W
• b.40 W
• c.100 W
• d.200 W✓

P = I^2 x R = 10^2 x 2 = 100 x 2 = 200 W. Trap: 20 W multiplies I x R, which gives the voltage drop (20 V), not power.

A motor delivers 10 HP of mechanical output while drawing 8,300 W of electrical input. What is its approximate efficiency?

• a.90%✓
• b.85%
• c.80%
• d.75%

10 HP = 7,460 W of output (746 W/HP). Efficiency = output / input = 7,460 / 8,300 = 0.90 = 90%. Trap: 75% forgets to convert HP to watts.

A load draws 60 kW of real power and 100 kVA of apparent power. What is the reactive power?

• a.60 kVAR
• b.40 kVAR
• c.80 kVAR✓
• d.100 kVAR

Reactive power = sqrt(kVA^2 - kW^2) = sqrt(100^2 - 60^2) = sqrt(6,400) = 80 kVAR.

A load has a current lagging the voltage by a phase angle of 30 degrees. What is its power factor?

• a.0.50
• b.0.87✓
• c.0.71
• d.0.94

Power factor = cos(theta) = cos(30 degrees) = 0.866, about 0.87.

What is the real power of a three-phase load at 480 V, 40 A, and 0.90 power factor?

• a.19,200 W
• b.24,000 W
• c.29,930 W✓
• d.33,254 W

P = 1.732 x V x I x PF = 1.732 x 480 x 40 x 0.90 = 29,930 W.

What is the real power of a single-phase load drawing 25 A at 240 V with a 0.85 power factor?

• a.6,000 W
• b.5,100 W✓
• c.7,200 W
• d.4,335 W

P = V x I x PF = 240 x 25 x 0.85 = 5,100 W.

In a balanced wye (star) three-phase system, the line-to-line voltage equals:

• a.2 times the line-to-neutral voltage
• b.0.577 times the line-to-neutral voltage
• c.1.732 times the line-to-neutral voltage✓
• d.the line-to-neutral voltage

In a wye system the line-to-line voltage is the square root of 3 (1.732) times the phase (line-to-neutral) voltage.

In a 240/120-V, three-phase, four-wire high-leg delta system, what is the voltage from the high (stinger) leg to the neutral?

• a.240 V
• b.277 V
• c.208 V✓
• d.120 V

The high leg to neutral in a 240-V delta is 240 x (sqrt(3)/2) = 208 V; that phase must never supply line-to-neutral loads.

An 80-kVA transformer serves a single-phase load at 240 V. What is the full-load current?

• a.333 A✓
• b.577 A
• c.240 A
• d.192 A

Single-phase current = VA / V = 80,000 / 240 = 333 A.

A 100-kW load at 0.70 power factor is corrected to 0.90 (tan of the 0.70 angle = 1.020, tan of the 0.90 angle = 0.484). How many kVAR of capacitance are required?

• a.20 kVAR
• b.48 kVAR
• c.100 kVAR
• d.53.6 kVAR✓

Qc = kW x (tan(theta1) - tan(theta2)) = 100 x (1.020 - 0.484) = 100 x 0.536 = 53.6 kVAR.

A transformer steps 480 V down to 240 V. What is its turns (voltage) ratio?

• a.8:1
• b.1:2
• c.2:1✓
• d.4:1

Turns ratio = primary voltage / secondary voltage = 480 / 240 = 2:1.

A transformer has a 10:1 turns ratio and a 5-A primary current. What is the ideal secondary current?

• a.5 A
• b.50 A✓
• c.15 A
• d.0.5 A

Current is inversely proportional to the turns ratio: Is = Ip x (Np/Ns) = 5 x 10 = 50 A.

A purely resistive load connected to 120 V draws 10 A. What is its resistance?

• a.130 ohms
• b.12 ohms✓
• c.1,200 ohms
• d.0.083 ohms

By Ohm's law, R = V / I = 120 / 10 = 12 ohms.

A resistive heating element rated 240 V has a resistance of 12 ohms. What power does it dissipate?

• a.28,800 W
• b.2,880 W
• c.20 W
• d.4,800 W✓

P = V^2 / R = 240^2 / 12 = 57,600 / 12 = 4,800 W.

A single-phase load draws 50 A at 240 V. What is the apparent power?

• a.12 kVA✓
• b.24 kVA
• c.6 kVA
• d.12 kW

Apparent power = V x I = 240 x 50 = 12,000 VA = 12 kVA (kW would require the power factor).

A motor delivers 25 HP of mechanical output. Ignoring losses, how many watts of output is that?

• a.25,000 W
• b.746 W
• c.33,540 W
• d.18,650 W✓

1 HP = 746 W, so 25 HP = 25 x 746 = 18,650 W of mechanical output.

A motor delivers 20 HP of mechanical output at 92% efficiency. What is its electrical input power?

• a.14,920 W
• b.13,726 W
• c.20,000 W
• d.16,217 W✓

Output = 20 x 746 = 14,920 W. Input = output / efficiency = 14,920 / 0.92 = 16,217 W.

A feeder conductor carries 20 A and has a total resistance of 0.5 ohm. What is the power lost as heat in the conductor?

• a.2,000 W
• b.200 W✓
• c.10 W
• d.40 W

Line loss = I^2 x R = 20^2 x 0.5 = 400 x 0.5 = 200 W.

What is the apparent power of an 80-kW load operating at a 0.80 power factor?

• a.80 kVA
• b.64 kVA
• c.125 kVA
• d.100 kVA✓

Apparent power = real power / power factor = 80 / 0.80 = 100 kVA.

For a 120-V rms sinusoidal supply, what is the approximate peak voltage?

• a.120 V
• b.170 V✓
• c.240 V
• d.85 V

Peak = rms x 1.414: 120 x 1.414 = 169.7 V, about 170 V.

In a three-phase, four-wire wye system with heavy nonlinear (electronic) load, why can the neutral current exceed the phase current?

• a.the neutral carries the vector sum of balanced fundamentals
• b.reactive current cancels in the neutral
• c.third-harmonic (triplen) currents are in phase across all three legs and add in the neutral✓
• d.the neutral is undersized by code

Triplen harmonics, principally the 3rd, are in phase across all three phases and add arithmetically in the neutral rather than canceling, so the neutral can carry more than a phase conductor.

A balanced three-phase panel has phase loads of A = 48 A, B = 36 A, C = 24 A. What is the ideal per-phase current if the loads were perfectly balanced?

• a.36 A✓
• b.24 A
• c.40 A
• d.48 A

Total = 48 + 36 + 24 = 108 A; perfectly balanced = 108 / 3 = 36 A per phase.

What three-phase transformer kVA rating corresponds to a full-load current of 200 A at 240 V?

• a.48 kVA
• b.100 kVA
• c.83 kVA✓
• d.144 kVA

kVA = (1.732 x V x I) / 1000 = (1.732 x 240 x 200) / 1000 = 83.1 kVA.

A 60-Hz sinusoidal waveform has a period of approximately:

• a.1.0 s
• b.8.3 ms
• c.60 ms
• d.16.7 ms✓

Period = 1 / frequency = 1 / 60 = 0.0167 s = 16.7 ms.