Question 12024 of 790 · California General Electrician Exam
Under the NEC 2020/2023 standard method, what is the demand load for general lighting, two small-appliance circuits and one laundry circuit in a 4,500 ft2 dwelling?
Exactly one answer is correct.
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Correct answer to question 12024
- D: 8,250 VA
Explanation
4,500 ft2 x 3 VA = 13,500 VA, plus 4,500 VA for the small-appliance and laundry circuits = 18,000 VA. First 3,000 VA at 100 % and the remaining 15,000 VA at 35 % gives 8,250 VA (Table 220.42 in 2020, Table 220.45 in 2023).
More questions: Load Calculations
- 12025Optional method (NEC 2020/2023): a 2,000 ft2 dwelling has two small-appliance circuits, one laundry circuit, a 12 kW range, a 5 kW dryer, a 1,200 VA dishwasher and a 4,500 W water heater. What is the calculated general load (heating and AC excluded)?
- 12026Optional method (NEC 2020/2023), 120/240 V service: 1,800 ft2 dwelling, two small-appliance and one laundry circuit, 11 kW range, 5.5 kW dryer, 1,000 VA dishwasher, 800 VA disposal, 4 kW water heater, 6,000 VA central AC and 8 kW central electric furnace. What is the calculated load in amperes?
- 12027Optional method (NEC 2020/2023): a 1,500 ft2 dwelling with two small-appliance and one laundry circuit, a 10 kW range, 5 kW dryer and 4.5 kW water heater has six separately controlled 2 kW baseboard heaters and a 4,000 VA AC unit. What is the total calculated load?
- 12032A dwelling has one electric dryer rated 3,800 W. What load must be used for it in the NEC 2020/2023 service calculation?
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