What is the difference between connected load and demand load?

B. Connected load adds everything at full rating; demand load applies factors because not everything runs at full load at once

A. They are the same number in different units

C. Demand load is always larger than connected load

D. Connected load only counts motors

Your load calculation has a demand factor applied to the range, and later you apply a general demand factor to the whole total as well. What has gone wrong?

B. The range has probably had demand applied twice; each load gets the factor the method gives it, once

A. Nothing — more factors give a smaller, better number

C. Ranges never get a demand factor

D. The general factor should have been applied first

A house has both electric heat and air conditioning. Why do many load methods not simply add both?

B. Because they are unlikely to run at the same time, so the method usually counts the larger (or treats them as non-coincident)

A. Because air conditioning is not an electrical load

C. Because heat is always ignored

D. Because they are on different services

Cuarto año Lesson 8 of 12 · about 16 min

Services and the load calculation method

What you’ll learn

The difference between connected load and demand load, and why the code lets you use demand.
The steps of a load calculation in order: area, basic load, demand, other loads, total, current.
How to apply stepped demand tiers without the classic mistakes.
How to turn total VA into amps for single-phase and three-phase services.

Before this: Three-phase power and power factor · Reading a panel schedule

A load calculation estimates the most current a building will realistically draw at once, so the service and feeders can be sized for it. It is a method, not a formula: a fixed set of steps, each fed with numbers from your code book. The steps are the same under CSA C22.1 and the NEC; the numbers are not. This lesson teaches the steps with invented figures — practice numbers, not code values — so you can practise the arithmetic and then plug in the real ones from your book.

Connected load and demand load

Add up the nameplate of everything in a house — range, dryer, heat, water heater, lighting, receptacles, the car charger — and you get the connected load. Nobody ever runs all of it at full power at the same moment. The code recognises that with demand factors: rules that let certain loads, or slices of a load, count at less than 100%. The result is the demand load, and it is what the service is sized from.

The method in six steps. Steps 2, 3 and 4 take their numbers from your code book; steps 5 and 6 are arithmetic.

The steps

Floor area. Measure it the way your book defines it — which floors count, whether you use inside or outside dimensions, how basements are treated.
Basic load. Some methods give a load per square foot or square metre; others give a fixed load for a first block of area and a smaller load for each additional block (sometimes with part of a block counting as a whole one).
Demand on the basic load. The book may let the first slice count in full and later slices at a lower percentage. Apply each percentage only to its own slice.
Other loads. Ranges, space heating, air conditioning, water heaters, dryers, EV chargers, pools and hot tubs — each with the factor the book gives it. Some depend on each other: heating and cooling are usually non-coincident.
Total demand. Add the basic load after demand and every other load after its factor.
Current. Single-phase: VA ÷ V. Three-phase: VA ÷ (√3 × V line-to-line). Then compare with the minimums and standard sizes your book sets — a separate step.

Worked: a house, with practice numbers (not code values)

Floor area 180 m² (about 1,940 ft²). Practice basic load: the first 100 m² = 4,500 VA, plus 1,200 VA for each additional 50 m² or part of 50 m².
Area beyond the first block: 180 − 100 = 80 m² = 1.6 blocks, and a part counts as whole → 2 blocks. Basic load = 4,500 + 2 × 1,200 = 6,900 VA.
Practice demand tiers: first 5,000 VA at 100%, the rest at 40%. 5,000 + (1,900 × 0.40 = 760) = 5,760 VA.
Range 12,000 VA at a practice 50% = 6,000 VA. Electric heat 10,000 VA and air conditioning 4,000 VA are non-coincident: count the larger, 10,000 VA. Water heater 3,000 VA at 100%. EV charger 7,200 VA at 100%.
Total demand = 5,760 + 6,000 + 10,000 + 3,000 + 7,200 = 31,960 VA.
At 240 V single-phase: 31,960 ÷ 240 = 133.2 A. That is the calculated load; the service rating that follows from it is a code-book step.

The same total on a three-phase supply

If that 31,960 VA were a small commercial load on 208 V three-phase: 31,960 ÷ (1.732 × 208) = 31,960 ÷ 360.3 = 88.7 A per line — assuming it is spread evenly across the phases.

The load calculation helper runs exactly these steps with the factors you type in, in square feet or square metres, and shows its working so you can check each line against your own.

En la obra

Tips

Lay the calculation out on paper in the same order every time, one line per step, units on every number.
Tab the load-calculation rules in your book together with the demand tables they point to.
Write the factor’s source next to each line (“basic load — from the book’s dwelling method”) so you can check it in a hurry.
Sanity-check the answer: a typical house lands in a band you will quickly get a feel for. A result wildly outside it usually means a missed zero or a factor applied twice.

Common mistakes

Applying the lower percentage to the whole load instead of only the slice above the first tier.
Applying demand twice — once to a load, again to the total.
Adding heating and cooling together when the method treats them as non-coincident.
Dividing a single-phase total by 208 or by √3 × 240.
Treating the calculated current as the service size, and skipping the book’s minimums.

Check yourself

0 right of 0 answered 7 questions · pick one answer each

1.What is the difference between connected load and demand load?
2.Using practice numbers (not code values): a basic load of 8 VA per unit area on a 150-unit floor. What is the basic load?
3.Practice tiers (not code values): the first 4,000 VA counts at 100% and the rest at 30%. The basic load is 10,000 VA. What is it after demand?
4.A total calculated demand of 36,000 VA on a 240 V single-phase service works out to how much current?
5.The same 36,000 VA of demand on a 208 V three-phase service is how much current per line?
6.Your load calculation has a demand factor applied to the range, and later you apply a general demand factor to the whole total as well. What has gone wrong?
7.A house has both electric heat and air conditioning. Why do many load methods not simply add both?

Answers

B. Connected load adds everything at full rating; demand load applies factors because not everything runs at full load at once — Connected load assumes every load runs flat out together, which never happens. Demand load applies the factors your code book allows to get a realistic figure. Demand is equal to or smaller than connected, never larger.
B. 1,200 VA — 150 × 8 = 1,200 VA. The method is always area × the load per unit area your book gives — the arithmetic is the easy part; getting the right factor and the right area definition is where marks are lost.
B. 5,800 VA — First 4,000 at 100% = 4,000. The remaining 6,000 at 30% = 1,800. Total 4,000 + 1,800 = 5,800 VA. The common mistakes are applying 30% to the whole 10,000 (3,000) or forgetting the remainder.
B. 150 A — Single-phase: I = VA ÷ V = 36,000 ÷ 240 = 150 A. Dividing by 208 or putting a √3 in would be the three-phase mistake. What service size that calls for is a separate step governed by your code book’s minimums and standard sizes.
B. 100 A — Three-phase: I = VA ÷ (√3 × V) = 36,000 ÷ (1.732 × 208) = 36,000 ÷ 360.3 ≈ 100 A. The √3 appears because the load is shared across three lines.
B. The range has probably had demand applied twice; each load gets the factor the method gives it, once — Double-dipping demand is one of the classic load-calc errors. Follow the method in your book step by step: each load gets its own factor once, in the place the method applies it, and the pieces are then added.
B. Because they are unlikely to run at the same time, so the method usually counts the larger (or treats them as non-coincident) — Heating and cooling are non-coincident loads — you do not heat and cool the same house at once. Load methods recognise this; exactly how your book handles it is a rule to look up, not guess.

Preguntas frecuentes

Why doesn’t this lesson give the real demand factors?

Because they are code-book content, they differ between CSA C22.1 and the NEC and between occupancies, and they change with each edition. The method — the order of the steps and the arithmetic — is the same everywhere, and it is what you need to learn. The numbers come from the book you are examined on and the one your inspector enforces.

Is the calculated current the service size?

No. The calculation gives you a minimum demand current. Your code book then sets minimum service sizes for the occupancy, and you choose from standard equipment ratings at or above what is required. That choice, and the conductor sizing that goes with it, are separate steps with their own rules.

Do dwelling and commercial calculations work the same way?

The shape is the same — a basic load, demand on it, specific loads with their own factors, total, divide by voltage — but the basic load, the factors and which loads get special treatment differ. Commercial and industrial calculations also lean more on the actual connected equipment, including continuous loads and the largest motor.

Ver también

Load calculation helper — walks these steps with your own factors
Rusty on load calculations? — the method in ten steps
Leer el directorio de un tablero
Three-phase power and power factor
Available fault current — the other number a service needs

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