A dwelling unit has a total calculated general lighting and receptacle load of 10,000 VA. What is the minimum number of 15-ampere, 120-volt general-purpose branch circuits required to serve this load, excluding laundry and small-appliance circuits?
- A7 circuits
- B8 circuits
- C6 circuits
- D5 circuits
Show answer & explanationAnswer & explanation
Correct answer: C. 6 circuits
According to NEC 210.11(A), the minimum number of branch circuits is determined by dividing the total load by the ampere rating of the circuit. For general lighting and receptacles, 3 VA per square foot is typically used, but the total VA is given. A 15-ampere, 120-volt circuit has a capacity of 15 A * 120 V = 1800 VA. To find the minimum number of circuits, divide the total load by the circuit capacity: 10,000 VA / 1800 VA/circuit = 5.55 circuits. Since you cannot have a fraction of a circuit, round up to the next whole number, which is 6 circuits.
Why the other options are wrong
- A. 7 circuits is more than the minimum required.
- B. 8 circuits is significantly more than the minimum required.
- D. 5 circuits is too few; the calculation requires rounding up.
General-Purpose Branch Circuit Count
The minimum number of general-purpose branch circuits for a dwelling unit is determined by dividing the total general lighting and receptacle load (in VA) by the VA rating of each individual circuit, then rounding up to the next whole number.
- General lighting/receptacles are calculated at 3 VA per square foot or given total VA.
- Circuit VA = Amperes * Volts.
- Always round up to the next whole number of circuits.
Memory trick: Divide total VA by circuit VA, then always round up for safety.