Power on set: 13A, 16A, 32A and load maths
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Power on set: 13A, 16A, 32A and load maths

Lights draw more than people expect, and an outlet rated for its maximum is not a working supply. Add up the watts, turn them into amps and compare them with the connector, then leave a margin.

In short: Amps equal total watts divided by volts. On a UK 230 V supply a 13 A plug tops out near 3 kW, a 16 A blue connector near 3.7 kW and a 32 A blue near 7.4 kW. A 32 A red three-phase connector at 400 V is near 22 kW. These are ceilings, not working loads, and a competent person must assess each supply.

What the sums are for

They tell you whether a supply can run a set before the cables go down. Add up the watts of every fixture, the camera, the monitors and the desk, then turn the total into amps and compare it with the connector. The result shows whether one supply is enough or whether the load must be split.

General guidance only. A competent person must assess every supply.

This tool does the sums. It does not check cable length, protection, how many outlets share a feed or the venue's actual supply. Those need a competent person on the day, working to BS 7909. Never run lighting from a domestic socket.

Supply

The planning limit is a margin used by this tool for loads that stay on for hours. It is not a figure taken from BS 7909 or BS 1363. Change it if your competent person gives you a different one.

Fixtures and equipment on this supply

Enter the rated watts from the label. For items rated in volt-amps (VA), enter the VA figure, which is the safer number. Include the camera, monitors, chargers and the desk, not only the lamps.

Result

Total load-
Current-
Share of rating-
0%100% of rating

How this is worked out
  • Total watts = the sum of (watts x quantity) for every row.
  • Amps = total watts / volts. This tool uses 230 V, the UK nominal single-phase supply. It treats watts as volt-amps, which is exact for tungsten lamps and slightly optimistic for some LED drivers and power supplies with a poor power factor.
  • Share of rating = amps / the connector's rating x 100. Ratings used: 13 A, 16 A, 32 A and 63 A.
  • Power at the rating = rating x 230 V, for example 13 x 230 = 2,990 W and 16 x 230 = 3,680 W.
  • The warning shows when the share is above the planning limit (80% unless you change it). Above 100% the supply is overloaded.
  • Three-phase supplies (for example the 32 A red connector at 400 V) use a different formula and are not covered here.

Connector ratings: BS 1363 and IEC 60309. Temporary systems at entertainment events: BS 7909.

The ratings

ConnectorSupplyRatingPower at the rating
13 A plug and socket (BS 1363)Single phase, 230 V13 AAbout 3 kW (13 × 230 = 2,990 W)
16 A blue, 2-pole and earthSingle phase, 230 V16 AAbout 3.7 kW (16 × 230 = 3,680 W)
32 A blue, single phaseSingle phase, 230 V32 AAbout 7.4 kW (32 × 230 = 7,360 W)
32 A red, three-phase with neutral and earthThree phase, 400 V between phases32 AAbout 22 kW (√3 × 400 × 32 = 22,170 W)

The red connector also gives 230 V between any phase and neutral, which the standard notes is available from that connector.

The calculator works on single-phase 230 V supplies only: 13 A (BS 1363), and 16 A, 32 A and 63 A blue connectors. A 63 A feed is about 14.5 kW at the rating (63 × 230 = 14,490 W). It has no three-phase option, so use the three-phase formula below by hand. It adds up your fixtures, shows the current and the percentage of the connector's rating, and warns when the load passes a planning limit, set at 80 per cent of the rating by default. That limit is the tool's own margin for loads that stay on for hours, not a figure from BS 7909 or BS 1363, and you can change it if your competent person gives you another.

Working out the current

For a single-phase load, amps = watts ÷ volts. Take five 1 kW lamps on a 230 V supply:

  • Total: 5 × 1,000 = 5,000 W.
  • Current: 5,000 ÷ 230 = about 21.7 A.
  • A single 16 A outlet cannot carry that, so the load needs a 32 A feed or two separate 16 A circuits.

For three-phase, power = √3 × line voltage × current. The sums change with the supply, so check the phase and voltage on the outlet before you use the formula.

Limits

These are the connector ratings, not the safe working loads. Cable length, the number of outlets on one feed, the protection on the supply and how the cable is laid all change what is safe. The fuse in a plug protects the cable and the plug, not the load you plan to run from them, so a fused plug does not make the supply suitable for the load.

A domestic socket shares its circuit with everything else on it. Do not run lighting from one, and plan a supply for tungsten heads and any large load. Do not treat an extension lead as a distribution system, ask whether the supply has RCD protection, and plan spare capacity on every circuit.

Common mistakes

  • Adding the watts for one outlet and forgetting the other outlets on the same feed.
  • Using the rating as the working load. A 16 A outlet with 16 A on it has no margin.
  • Mixing up the 230 V and 400 V supplies, so the sums give the wrong current.
  • Assuming the venue's supply is free to use. Confirm what it can supply before the day.

Safety

This is general guidance. A competent person must assess the actual supply, cables, distribution and loads for each job. BS 7909 is the code of practice for temporary low-voltage a.c. systems at entertainment events, and it is written for the people responsible for safety, including producers, hire companies and electrical contractors. Follow it, and the competent person's assessment, rather than this page.

Where it fits

The fixtures on the lighting plot set the watts. Before you draw the plot, ask what the room can supply. The studio hire page explains what each space includes.

Sources

Shooting this for real? Book the room and the crew as one job. We will tell you honestly which roles your shoot needs. Hire the rooms, add crew →

Related guides

Questions we get asked

How many watts can a 13 A plug carry?

About 3 kW at 230 V (13 A × 230 V is 2,990 W). Treat that as the ceiling for the outlet, not a working load, and have a competent person assess it.

How do I work out the current for a lighting load?

Divide the total watts by the supply voltage. Five 1 kW lamps on 230 V draw about 21.7 A, more than a single 16 A outlet can safely carry.

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