Free NEC Code Questions and Answers

Question 1

A set of 450 kcmil feeder conductors is protected by a 400 breaker. 
With a 25 A over current device, a feeder fed from a 450 kcmil feeder
 disconnects. What is the smallest conductor size for these taps?

Accepted Answer

8 AWG

Answer

This scenario likely involves a continuous load calculation for the tap conductor. If the 25A overcurrent device is protecting a continuous load, the conductor must be sized for 125% of that load, which is 25A * 1.25 = 31.25A. According to NEC Table 310.16 (75°C column), 10 AWG copper is rated for 30A (insufficient), while 8 AWG copper is rated for 50A, making it the smallest suitable conductor.

Question 2

For a 4in * 4 in wireway, what is the maximum permitted conductor fill?

Accepted Answer

3.2 sq in

Answer

According to NEC Article 376.22(A), the sum of the cross-sectional areas of all conductors in a wireway shall not exceed 20% of the wireway's interior cross-sectional area. For a 4-inch by 4-inch wireway, the interior area is 16 square inches. Therefore, the maximum permitted conductor fill is 20% of 16 sq in, which equals 3.2 sq in.

Question 3

For a 4000 W, 240 electric water heater, what size branch circuit conductor is required?

Accepted Answer

10 AWG

Answer

First, calculate the current: I = P/V = 4000W / 240V = 16.67A. Since a water heater is a continuous load, the conductor must be sized for 125% of the load: 16.67A * 1.25 = 20.84A. According to NEC Table 310.16 (75°C column), 12 AWG copper is rated for 25A, but NEC 240.4(D) limits its overcurrent protection to 20A. Therefore, 10 AWG copper, rated for 30A, is required to safely carry 20.84A and be protected by a standard overcurrent device (e.g., 25A or 30A).

Question 4

What is the general lighting load of a hospital with an area of 80,000 sq ft after the demand factor is applied?

Accepted Answer

42,000 VA

Answer

To determine the general lighting load, first calculate the total VA before demand: 80,000 sq ft * 2 VA/sq ft (from NEC Table 220.12 for hospitals) = 160,000 VA. Then, apply the demand factors from NEC Table 220.42: the first 50,000 VA at 40% (20,000 VA) and the remaining 110,000 VA (160,000 - 50,000) at 20% (22,000 VA). Summing these gives a total demand load of 42,000 VA.

Question 5

For a dwelling unit with a calculated load of 240 A, what size feeder conductors 
are required? The feeder conductors carry the whole load of the dwelling unit, 
and the service disconnect is rated for 250 A?

Accepted Answer

4/0 AWG

Answer

For dwelling unit feeder conductors, NEC Table 310.12 provides specific conductor sizes. While the service disconnect is rated for 250A and the calculated load is 240A, 4/0 AWG is a standard size for a 200A service when using aluminum conductors. This size is often considered sufficient for dwelling unit loads up to 240A, especially when applying the special allowances for dwelling unit services and feeders.

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