FREE Journeyman Electrician’s Service and Feeder Calculations Questions and Answers

Question 1

Using the standard calculation method, what is the total service load for a 2,000 sq. ft. single-family dwelling with a 12 kW electric range and a 5 kW electric dryer?

Accepted Answer

18,625 VA

Answer

The calculation is as follows: General lighting (2,000 sq. ft. x 3 VA) = 6,000 VA. Small appliance circuits (2 x 1,500 VA) = 3,000 VA. Laundry circuit (1 x 1,500 VA) = 1,500 VA. Total general load = 10,500 VA. Applying demand factors from NEC Table 220.42: First 3,000 VA at 100% = 3,000 VA. Remainder (7,500 VA) at 35% = 2,625 VA. Net general load = 5,625 VA. The range demand is 8,000 VA (from Table 220.55, Column C for one 12 kW range). The dryer load is 5,000 VA (minimum per 220.54). Total service load = 5,625 VA + 8,000 VA + 5,000 VA = 18,625 VA.

Question 2

A 120/240V feeder supplies a non-continuous load of 10,000 VA from line-to-neutral lighting and receptacles, plus one 14 kW household electric range. What is the calculated neutral load for this feeder?

Accepted Answer

16,160 VA

Answer

The neutral carries the full 10,000 VA of line-to-neutral load. For the range, first calculate the demand load from Table 220.55. A 14 kW range is 2 kW over 12 kW. The demand is 8 kW plus 5% for each kW over 12: 8,000 VA + 2*(8,000 VA * 0.05) = 8,800 VA. According to NEC 220.61(B), the neutral load for a household range can be calculated at 70% of the demand load: 8,800 VA * 0.70 = 6,160 VA. The total neutral load is the sum: 10,000 VA + 6,160 VA = 16,160 VA.

Question 3

A feeder in a commercial kitchen supplies five pieces of thermostatically controlled kitchen equipment with a total connected load of 41 kW. What is the calculated demand load for the feeder?

Accepted Answer

28.7 kW

Answer

According to NEC Table 220.56, Demand Factors for Commercial Electric Cooking Equipment, the demand factor for 5 units is 70%. The calculated demand load is the total connected load multiplied by the demand factor: 41 kW * 0.70 = 28.7 kW.

Question 4

A new single-family dwelling is supplied by a 200A, 120/240V single-phase service. What is the minimum size for the ungrounded copper service-entrance conductors with THWN insulation permitted by the NEC?

Accepted Answer

2/0 AWG

Answer

NEC 310.12(A) permits service conductors for a 100A through 400A single-family dwelling service to have an ampacity of not less than 83% of the service rating. For a 200A service, the minimum conductor ampacity is 200A * 0.83 = 166A. Referring to NEC Table 310.16, using the 75°C column for copper conductors, a 2/0 AWG conductor has an ampacity of 175A, which is the smallest size rated for at least 166A.

Question 5

A feeder supplies 15 individual dwelling units in an apartment building. The calculated load for each unit is 14 kVA. Using the optional method, what is the calculated demand load for the feeder?

Accepted Answer

88.2 kVA

Answer

First, find the total connected load: 15 units * 14 kVA/unit = 210 kVA. Then, refer to NEC Table 220.84, Optional Calculation—Demand Factors for Three or More Multifamily Dwelling Units. For 15 dwelling units, the demand factor is 42%. The calculated demand load is 210 kVA * 0.42 = 88.2 kVA.

Journeyman Electrician’s Exam Practice Test

Journeyman Electrician’s Exam Practice Test

FREE Journeyman Electrician’s Service and Feeder Calculations Questions and Answers