Circuit calculations

[HarryDanielsLD]

Okay,

So I'm relatively new to lighting, I've been helping out at my local theatre for a while but still dont know a lot and have to pick up a lot of knowledge on the job.

I was just wondering how to work out how many lights you can place on a circuit?

I know the basic calculation W divided by V = A
Am I to do this calculation for each lantern and make sure the amps dont exceed the amps in the circuit?

I'm slightly confused and would appreciate any help!

Thanks

 

[xander]

If you are running everything at the same voltage, you can cut out the middle step. For instnace, here in America we use nominally 120V and use 12AWG cable with a 20A circuit breaker. This means we can safely put 120V * 20A = 2400W on each circuit. Just add up your lamps and make sure they stay under 2400W. Which is generally 4 Source4s when lamped at 575W.

-Tim

 

[SteveB]

As additional to Xander's rely, be aware that should you be running the circuit for more then 3 hours continuously, you need to down rate the load to 80% of the circuit breaker rating, unless you are using a dimmer or circuit designed for continuous loading. ETC Sensor dimmers, as example are designed for continuous loads and can safely run their circuits at the breaker rating. Thus in Xander's example, 4x575 watt Source 4's are OK on a Sensor 2400 watt dimmer. Other dimming systems may not be rated for continuous use. On a non-rated 2400 watt circuit, the load can only be 1920 watts.


Sent from my iPad using Tapatalk

 

[JD]

On US power, the "divide by 100 and round up" rule gives you a fast ballpark as you only have to move the decimal point. The exact number is of course wattage divided by RATED voltage. In other words, if a 1k lamp is rated at 120 volts, and is running at 120 volts, then 1000/120 = 8.33 amps. The "divide" rule would say 1000.00 (move decimal point) 10.00 amps, building a bit of a safety in. Your 120 volt 575 watt HPL running at 120 volts - 575 (move dp) 5.75 round up to 6 amps. That actual builds in your full 80% derate for normal breakers! (575/120 = 4.79, 6 x 80% = 4.8)

Now, here's one thing to watch out for- a 1k 115 volt lamp running at 120 volts will draw more power than a 1k 120 volt lamp!
The 115 volt lamp has a hot resistance of 13.2 ohms. The 120 volt lamp has a hot resistance of 14.41 ohms. In other words, besides shortening the life, the 115 lamp would be drawing almost 1100 watts. (Exception- voltage regulated dimmers.) Of course, dimmers eat up a couple of volts, but you have to keep these things in mind if you live in an area where line voltage is more like 125 to 127 volts!

One last note- Filaments have different resistances at different temperatures, so do not assume that lowering your lamp to 50% will reduce it's current draw 50%, even on a liner (no curve) dimmer.

 

[STEVETERRY]

Okay,

So I'm relatively new to lighting, I've been helping out at my local theatre for a while but still dont know a lot and have to pick up a lot of knowledge on the job.

I was just wondering how to work out how many lights you can place on a circuit?

I know the basic calculation W divided by V = A
Am I to do this calculation for each lantern and make sure the amps dont exceed the amps in the circuit?

I'm slightly confused and would appreciate any help!

Thanks

All of the previous discussion works for a purely resistive load like a tungsten lamp. For an arc source or LED fixture, the power factor of the load needs to be considered.

In the case of a fixture with a switch mode power supply, you need to think in amps (I) not watts (W). Use the formula:

I = P/E
-----pf

Where I = current in amps, pf = Power factor of the load, E= voltage., and P = Power in watts

For example:

1000 watt load at 120V with a Power factor of 0.8:


1000/120 = 8.33A
8.33/ 0.8 = 10.41 A

In this case, you need to size the circuit and overcurrent protective device for 10.41A of load, not the traditional 1000W (8.33A) of load.

Many non-linear loads using switch mode power supplies have a Power Factor (pf) far less than one. Therefore, you cannot "fudge" the calculation using the "old" formula of I=P/V , especially as our rigs move to more and more non-linear loads and less tungsten loads.

ST