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A dwelling has 1,800 square feet of floor area. Calculate the minimum general lighting load in volt-amperes (VA) before applying demand factors. Identify the NEC reference.
Answer: 5,400 VA
Calculation: \(1,800\text{ ft}^2 \times 3\text{ VA/ft}^2 = 5,400\text{ VA}\).
Code: 220.12 and Table 220.12.
Explanation: The dwelling-unit general lighting load is calculated using 3 VA per square foot.
What is the minimum calculated load, in VA, for the two required small-appliance branch circuits in a dwelling? State the minimum load assigned to each circuit and give the code reference.
Answer: 3,000 VA total; 1,500 VA per circuit.
Calculation: \(2 \times 1,500 = 3,000\text{ VA}\).
Code: 210.11(C)(1) and 220.52(A).
Explanation: A dwelling requires at least two small-appliance branch circuits, and each is assigned a 1,500 VA load for the calculation.
What minimum load must be included in a dwelling-unit calculation for the required laundry branch circuit? Give the NEC reference.
Answer: 1,500 VA.
Code: 210.11(C)(2) and 220.52(B).
Explanation: At least one laundry branch circuit is required, and 1,500 VA is included in the dwelling load calculation.
A dwelling has a calculated general lighting load of 5,400 VA, two required small-appliance circuits totaling 3,000 VA, and a laundry load of 1,500 VA. Using the standard-method demand factors for dwelling general lighting and the applicable included loads, calculate the demand load. Show your work and cite the NEC table.
Answer: 5,415 VA.
Total applicable load: \(5,400+3,000+1,500=9,900\text{ VA}\).
First 3,000 VA at 100%: 3,000 VA.
Remaining 6,900 VA at 35%: \(6,900\times0.35=2,415\text{ VA}\).
Total demand: \(3,000+2,415=\boxed{5,415\text{ VA}}\).
Code: 220.42 and Table 220.42; see also 220.52(A) and (B).
Explanation: The qualifying dwelling lighting, small-appliance and laundry loads are combined and subjected to the applicable demand factors.
A dwelling has four qualifying appliances fastened in place, each rated 1,200 VA. Assume the appliances meet the conditions for the demand factor in the NEC. Calculate the combined demand load.
Answer: 3,600 VA.
Calculation: \(4\times1,200=4,800\text{ VA}\); \(4,800\times0.75=3,600\text{ VA}\).
Code: 220.53.
Explanation: A 75% demand factor is permitted for four or more qualifying appliances fastened in place, subject to the section's conditions and exclusions.
A household electric dryer has a nameplate rating of 5,600 W. What load must be used for one dryer in the dwelling calculation? Explain why the nameplate rating matters and cite the NEC section.
Answer: 5,600 W (5.6 kW).
Code: 220.54.
Explanation: One household dryer is calculated at 5,000 W or its nameplate rating, whichever is larger. Since the nameplate is 5,600 W, use 5,600 W.
A 240-volt water heater is rated at 4,500 W. Calculate its rated current in amperes. Show the formula.
Answer: 18.75 A.
Calculation: \(I=P/V=4,500/240=\boxed{18.75\text{ A}}\).
Code: 220.5(B) for calculation rules; equipment and circuit requirements depend on the installation.
Explanation: For a resistive load, watts divided by volts gives current.
A 240-volt, single-phase load draws 12,000 W. Calculate the current, assuming a power factor of 1.0.
Answer: 50 A.
A 120-volt continuous load draws 16 A. Calculate the minimum conductor ampacity and the minimum standard overcurrent-device rating based on the 125% sizing rule, assuming no other adjustment or correction factors apply.
Answer: 20 A minimum ampacity and a 20 A standard overcurrent-device rating, subject to applicable conductor and equipment requirements.
Calculation: \(16\times1.25=20\text{ A}\).
Code: 210.19(A)(1), 210.20(A), and 240.6(A).
Explanation: Branch-circuit conductors and overcurrent protection must account for 125% of the continuous load under the applicable rules. Final conductor selection must also account for terminal temperature ratings and any adjustment or correction factors.
A building has electric space heating and air conditioning that cannot operate simultaneously because of an interlocked control system. Explain how these loads are treated in a calculation and identify the applicable NEC section.
Answer: The larger of the two loads may be used when the loads are noncoincident under the applicable Code conditions.
Code: 220.60.
Explanation: Loads that will not operate simultaneously need not necessarily be added together. The stated interlock supports the noncoincident-load treatment.
Under the dwelling-unit standard calculation method, explain how the demand factor applies to the first 3,000 VA and the portion above 3,000 VA of the applicable general lighting load. Identify the applicable NEC table.
Answer: The first 3,000 VA is calculated at 100%; the portion from 3,001 VA through 120,000 VA is calculated at 35% for dwelling units.
Code: Table 220.42.
Explanation: Demand factors account for the fact that not all connected lighting and applicable receptacle loads are expected to operate at full load simultaneously.
A dwelling contains one household range rated at 12 kW. Explain how the calculated load is determined under the standard method. Identify the applicable table and explain why simply using the full nameplate rating may not be the correct calculated demand.
Answer: 8 kW calculated demand.
Code: Table 220.55, Column C, and Note 1 where applicable.
Explanation: For one household range rated 12 kW, Table 220.55 permits an 8 kW demand load under the standard method. The table's demand allowance is not necessarily the same as the appliance's nameplate rating.
Explain why the neutral conductor load for a dwelling service is not necessarily equal to the sum of all connected line-to-neutral loads. Identify the NEC section that addresses the calculated neutral load.
Answer: The calculated neutral load can be less than the sum of all connected line-to-neutral loads because the applicable calculation rules account for load diversity and the relationship between loads on opposite legs.
Code: 220.61.
Identify the NEC section that provides the optional method for calculating the service load for a qualifying dwelling unit. State one important condition that must be met before this method can be used.
Answer: NEC 220.82, Optional Calculation — Dwelling Unit.
Explanation: This method is available only when the dwelling and service or feeder meet the section's eligibility conditions, including the applicable minimum rating and permitted system configuration. It is not automatically available for every building.
Explain the difference between watts and volt-amperes. Why can a circuit's current calculation require power factor information when the load is rated in watts?
Answer: Watts measure real power; volt-amperes measure apparent power. For AC loads, \(P=VI\times PF\) for a single-phase load under the usual sinusoidal definition of power factor.
Explanation: When power factor is below 1.0, the current required for a given wattage can be higher than the current calculated by simply dividing watts by volts.
A 20 A load is supplied by #12 AWG copper conductors over a 100-foot one-way run. Calculate the voltage drop in volts and as a percentage of the nominal circuit voltage.
Approximately 7.90 V, or 6.58%.
The calculated drop is above the NEC's general 3% branch-circuit recommendation.
A 16 A load is supplied by #10 AWG copper conductors over a 75-foot one-way run on a 120 V circuit. Calculate the voltage drop and percentage.
Approximately 2.98 V, or 2.48%.
This is below the general 3% branch-circuit recommendation.
A 20 A load is supplied by #10 AWG copper conductors over a 150-foot one-way run on a 240 V single-phase circuit. Calculate the voltage drop and percentage.
Approximately 7.46 V, or 3.11%.
The calculated drop slightly exceeds the general 3% branch-circuit recommendation
A 15 A load operates on a 120 V circuit supplied by #12 AWG copper conductors over a 50-foot one-way run. Calculate the voltage drop and percentage.
Approximately 2.96 V, or 2.47%
A balanced 208 V, three-phase load draws 20 A. The circuit uses #12 AWG copper conductors over a 100-foot one-way run. Calculate the voltage drop in volts and as a percentage of 208 V.
Approximately 6.84 V, or 3.29%.
Use the balanced three-phase formula, not the single-phase factor-of-two formula.