NEC Load Calculation Practice Problems: 10 Exam-Style Questions with Step-by-Step Solutions

Load calculations are the single biggest topic on the Texas Master Electrician exam. They're also the #1 reason people fail.

The good news? Load calculations follow a repeatable process. Once you learn the method and practice enough problems, they become points you can count on.

This post gives you 10 exam-style practice problems. Each one includes the full step-by-step solution with NEC article references. Work through them with your codebook open — that's exactly how you'll do it on exam day.

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Before You Start

You'll need your NEC codebook open to these sections:

• NEC 220.12 — Lighting load by occupancy type
• NEC 220.42 — General lighting demand factors
• NEC 220.44 — Receptacle load demand factors
• NEC 220.50 & 220.51 — Motor load calculations
• NEC 220.55 — Cooking equipment demand factors (Table 220.55)
• NEC 220.82 — Optional calculation for dwelling units
• NEC 430.52 — Motor branch-circuit short-circuit and ground-fault protection
• Table 310.16 — Conductor ampacities
• Chapter 9, Table 1 — Conduit fill percentages

Grab a calculator. Show your work. Don't peek at the answers until you've tried each problem.

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Problem 1: Residential General Lighting Load

A single-family dwelling has 2,400 square feet of living space. Calculate the general lighting load before applying demand factors.

Solution

Per NEC 220.12, Table 220.12, the unit load for a dwelling unit is 3 VA per square foot.

General lighting load = 2,400 sq ft × 3 VA/sq ft

General lighting load = 7,200 VA

This is the value before demand factors. We'll apply those in the next problem.

NEC Reference: Table 220.12

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Problem 2: Applying Demand Factors to General Lighting

Using the 7,200 VA general lighting load from Problem 1, plus two small-appliance circuits (1,500 VA each per NEC 220.52(A)) and one laundry circuit (1,500 VA per NEC 220.52(B)), calculate the demand load.

Solution

First, add up the total connected load:

Now apply the demand factors from NEC Table 220.42:

• First 3,000 VA at 100% = 3,000 VA
• Remaining 8,700 VA at 35% = 3,045 VA

Demand load = 3,000 + 3,045 = 6,045 VA

This is a significant reduction from 11,700 VA. The demand factors account for the fact that not everything runs at the same time.

NEC Reference: Table 220.42, 220.52(A), 220.52(B)

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Problem 3: Cooking Equipment Demand (Single Range)

A dwelling unit has one 12 kW household electric range. What is the demand load for this range?

Solution

Per NEC Table 220.55, Column C, for a single household range rated not more than 12 kW:

Maximum demand = 8 kW (8,000 VA)

This is a straight table lookup. For a single range 12 kW or less, Column C gives you 8 kW.

If the range were rated over 12 kW, you'd need to apply the notes below Table 220.55 to adjust the Column C value. That's a common exam trick — make sure you read the notes.

NEC Reference: Table 220.55, Column C

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Problem 4: Cooking Equipment Demand (Multiple Ranges)

An apartment building has 20 dwelling units, each with a 10 kW electric range. Calculate the total demand load for all ranges.

Solution

Per NEC Table 220.55, Column C, for 20 appliances:

Look across the row for 20 appliances in Column C.

Maximum demand = 35 kW

Compare this to the connected load: 20 × 10 kW = 200 kW. The demand factor brings it down to 35 kW — that's only 17.5% of the connected load. This is why demand factors matter so much in commercial calculations.

NEC Reference: Table 220.55, Column C

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Problem 5: Dryer Load Calculation

A dwelling has one 5.5 kW electric clothes dryer. What is the demand load?

Solution

Per NEC 220.54, the demand load for a single dryer shall not be less than 5,000 VA (5 kW) or the nameplate rating, whichever is larger.

Since 5,500 VA > 5,000 VA:

Demand load = 5,500 VA

If the dryer were rated at only 4 kW, you'd still use 5,000 VA as the minimum.

NEC Reference: NEC 220.54

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Problem 6: Motor Branch Circuit Conductor Sizing

A single 460V, 3-phase, 50 HP continuous-duty motor needs branch circuit conductors. The motor has a nameplate FLA of 62 amps. Determine the minimum conductor ampacity required.

Solution

Important: For motor circuit calculations, always use NEC Table 430.250 (full-load current tables), NOT the motor nameplate. This is a common exam trap.

Per NEC Table 430.250, a 50 HP, 460V, 3-phase motor has a full-load current of 65 amps.

Per NEC 430.22, branch circuit conductors must have an ampacity of not less than 125% of the motor full-load current:

Minimum conductor ampacity = 65 A × 1.25

Minimum conductor ampacity = 81.25 amps

Per NEC Table 310.16, using 75°C column with THWN-2 copper conductors:

A 4 AWG copper conductor is rated for 85 amps — this is the minimum size.

NEC Reference: Table 430.250, 430.22, Table 310.16

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Problem 7: Motor Overload Protection

Using the same 50 HP motor from Problem 6 (nameplate FLA of 62 amps, service factor 1.15), determine the maximum overload protection.

Solution

Per NEC 430.32(A)(1), for a motor with a service factor of 1.15 or greater, the overload device shall be rated at not more than 125% of the motor nameplate current.

Maximum overload = 62 A × 1.25

Maximum overload protection = 77.5 amps

Note: Here we DO use the nameplate current (62A), not the table value (65A). Overload protection is based on the actual motor, while conductor sizing is based on the table. The exam loves testing whether you know which value to use where.

If the service factor were less than 1.15, you'd use 115% instead of 125%.

NEC Reference: NEC 430.32(A)(1)

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Problem 8: Motor Short-Circuit Protection (Inverse-Time Breaker)

For the same 50 HP, 460V, 3-phase motor, determine the maximum size inverse-time circuit breaker for branch circuit short-circuit and ground-fault protection.

Solution

Per NEC 430.52 and Table 430.52, for an inverse-time circuit breaker, the maximum rating is 250% of the motor full-load current.

Again, use the NEC table value (65A), not the nameplate:

Maximum breaker size = 65 A × 2.50 = 162.5 A

Per NEC 430.52(C)(1), Exception No. 1, if 162.5A does not correspond to a standard breaker size, you may round UP to the next standard size.

Standard breaker sizes: 15, 20, 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150, 175, 200...

Maximum breaker size = 175 amps

NEC Reference: Table 430.52, 430.52(C)(1) Exception No. 1, 240.6(A)

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Problem 9: Optional Dwelling Calculation (NEC 220.82)

A 2,200 sq ft dwelling has the following loads:

• General lighting and receptacles (calculated per 220.12 and 220.52)
• One 12 kW range
• One 5.5 kW dryer
• One 4.5 kW water heater
• One 6 kW A/C unit (largest motor)
• One 10 kW electric heat (3 units at 3.33 kW each)

Use the optional method (NEC 220.82) to calculate the total demand load.

Solution

Step 1: Calculate the general loads per 220.82(B):

Step 2: Add all other loads at nameplate:

Step 3: Apply 220.82(B) demand factors:

Total of general + other loads = 11,100 + 22,000 = 33,100 VA

Per 220.82(B):

• First 10,000 VA at 100% = 10,000 VA
• Remaining 23,100 VA at 40% = 9,240 VA

Subtotal = 10,000 + 9,240 = 19,240 VA

Step 4: Heating vs. A/C — use the LARGER (NEC 220.82(C)):

• A/C: 6,000 VA (use 100%)
• Heat: 10,000 VA (use 40% per 220.82(C)(6) — but check your code year)

For the 2023 NEC, electric heat uses 40% for the optional method: Heat demand = 10,000 × 0.40 = 4,000 VA

A/C at 6,000 VA is larger, so use A/C.

Step 5: Total demand:

19,240 + 6,000 = 25,240 VA

Total demand load = 25,240 VA

At 240V single-phase: 25,240 ÷ 240 = 105.2 amps

NEC Reference: NEC 220.82, 220.82(B), 220.82(C)

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Problem 10: Conduit Fill Calculation

You need to install four 10 AWG THHN conductors and three 12 AWG THHN conductors in EMT conduit. What is the minimum trade size conduit required?

Solution

Step 1: Find the cross-sectional area of each conductor from NEC Chapter 9, Table 5:

Step 2: Determine the fill percentage from NEC Chapter 9, Table 1:

With 7 conductors (more than 2), the maximum fill is 40%.

Step 3: Calculate the minimum conduit area needed:

Minimum conduit area = 0.1243 ÷ 0.40 = 0.3108 sq in

Step 4: Find the conduit size from NEC Chapter 9, Table 4 (EMT):

0.3108 sq in > 0.304 sq in (1/2" is too small)

0.3108 sq in < 0.533 sq in (3/4" works)

Minimum conduit size = 3/4" EMT

NEC Reference: Chapter 9, Tables 1, 4, and 5

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Key Takeaways for Exam Day

Tab these tables in your codebook — you'll reference them constantly:

Common traps to watch for:

1. Using motor nameplate amps instead of NEC table values for conductor sizing and short-circuit protection

2. Forgetting to use nameplate amps for overload protection (the opposite of #1)

3. Not applying demand factors — the exam will give you connected loads and expect you to reduce them

4. Mixing up 220.82 optional method with the standard method — know when each applies

5. Forgetting the 125% multiplier for continuous loads

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Ready for More Practice?

These 10 problems cover the most heavily tested load calculation topics. If you got 7 or more right on your first try, you're in solid shape.

If you struggled, don't worry — that's exactly why you're practicing now instead of on exam day.

Next steps:

• Take our free practice quizzes covering all 6 exam topic areas
• Review the Formula Reference Sheet for quick calculation reminders
• Follow the 4-Week Study Plan for a structured approach
• Consider the Master Electrician Future Kit for 30+ premium practice questions with detailed explanations

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