Vermont Journeyman Electrician Calculations Practice

A 120 V source drives 15 A through a resistive load. What is the resistance of the load?

Resistance equals voltage divided by current.

When voltage steps down, current steps up by the same ratio.

A 208 V three-phase feeder carries 65 A over a one-way length of 200 ft using 3 AWG copper conductors (52,620 cmil). Use K = 12.9. What is the voltage drop?

On three-phase circuits √3 (1.732) replaces the 2 used for single-phase.

A single-phase resistive load draws 15 A at 240 V. How many watts does it use?

Power in watts equals volts multiplied by amperes for a resistive single-phase load.

A metal box contains 5 insulated 10 AWG conductors, 1 device (switch or receptacle) on a single yoke, 2 equipment grounding conductors and internal cable clamps. The volume allowance for each 10 AWG conductor is 2.5 in³ (given). What is the minimum box volume?

Count conductor volumes first (devices double, up to four equipment grounds once, clamps once), then multiply by the volume for the largest conductor.

A three-phase motor load draws 70 A per line at 480 V with a power factor of 0.8. What is the true power in kW?

True power on a three-phase system is √3 × volts × amperes × power factor.

A 480 V circuit supplies a resistive load of 14 Ω. What current flows in the circuit?

Current equals voltage divided by resistance.

What is the full-load secondary line current of a 225 kVA three-phase transformer with a 480 V secondary?

Three-phase line current divides the VA by √3 × line voltage.

A 480 V three-phase feeder carries 135 A over a one-way length of 390 ft using 4/0 AWG aluminum conductors (211,600 cmil). Use K = 21.2. What is the voltage drop?

On three-phase circuits √3 (1.732) replaces the 2 used for single-phase.

A single-phase resistive load draws 36 A at 120 V. How many watts does it use?

Power in watts equals volts multiplied by amperes for a resistive single-phase load.

A metal box contains 9 insulated 12 AWG conductors, 1 device (switch or receptacle) on a single yoke, 3 equipment grounding conductors and internal cable clamps. The volume allowance for each 12 AWG conductor is 2.25 in³ (given). What is the minimum box volume?

Count conductor volumes first (devices double, up to four equipment grounds once, clamps once), then multiply by the volume for the largest conductor.

A 75 kW three-phase load operates at 208 V with a power factor of 0.9. What is the line current?

Three-phase current uses √3 (1.732) and the power factor in the denominator. Leaving out either one is the most common mistake.

Three resistors of 30 Ω, 38 Ω and 14 Ω are connected in series. What is the total resistance?

In series the resistances simply add.

A branch circuit supplies 36 A of continuous load and 11 A of noncontinuous load. The overcurrent device is not listed for 100% continuous operation. What is the minimum overcurrent device rating, before rounding up to a standard size?

Only the continuous part (3 hours or more) is taken at 125%; the noncontinuous part counts at 100%.

A raceway has a total internal area of 1.496 in². Each 10 AWG conductor, including insulation, has an area of 0.0211 in² (both given). With more than two conductors, fill is limited to 40%. What is the maximum number of conductors allowed?

Multiply the total area by 40% and divide by one conductor’s area. Round down, unless the decimal is 0.8 or more: then Chapter 9, Note 7 rounds up.

Three resistors of 44 Ω, 29 Ω and 28 Ω are connected in parallel. What is the total resistance?

Add the reciprocals, then take the reciprocal of the sum. The result is smaller than the smallest resistor.

A branch circuit supplies 25 A of continuous load and 30 A of noncontinuous load. The overcurrent device is not listed for 100% continuous operation. What is the minimum overcurrent device rating, before rounding up to a standard size?

Only the continuous part (3 hours or more) is taken at 125%; the noncontinuous part counts at 100%.

A raceway has a total internal area of 3.356 in². Each 10 AWG conductor, including insulation, has an area of 0.0211 in² (both given). With more than two conductors, fill is limited to 40%. What is the maximum number of conductors allowed?

Multiply the total area by 40% and divide by one conductor’s area. Round down, unless the decimal is 0.8 or more: then Chapter 9, Note 7 rounds up.

The 2 accounts for current flowing out and back on a single-phase circuit; L is the one-way length.

The 2 accounts for current flowing out and back on a single-phase circuit; L is the one-way length.