[herdfan]

Ok, I am somewhat stumped by a problem surrounding this formula. So here is a scenario:

3 600W Rheostat dimmers on a 15A 14/2 circuit. On each dimmer are 9x65W can lights. So each dimmer has 585W and the 15A circuit has a total of 1755W. I am ignoring the 20% cushion for this example.

So 1755W/120V = 14.625A which is fine for the 15A Circuit with 14/2.

Now I dim the all 3 dimmers to 50%. Since it is a Rheostat dimmer and not a clipped wave dimmer, the output voltage would be 60V. Using the same formula:

1755W/60V = 29.25A

Why does this not over Amp the circuit? What am I missing?

The only thing I can think of is that the lights are no longer using 1755W.

Thanks

 

[Big_John]

Using the same formula:

1755W/60V = 29.25A
The only thing I can think of is that the lights are no longer using 1755W.

That's exactly it.

Light produced is directly related to energy consumed. If you're dimming the lights you are lowering the voltage which reduces the amount of power each light uses.

It's a bit more complicated than this, but the simple math is at 60V you're drawing somewhere in the neighborhood of 7.5A and producing 450 watts.

Ok, I am somewhat stumped by a problem surrounding this formula. So here is a scenario:

3 600W Rheostat dimmers on a 15A 14/2 circuit. On each dimmer are 9x65W can lights. So each dimmer has 585W and the 15A circuit has a total of 1755W. I am ignoring the 20% cushion for this example.

So 1755W/120V = 14.625A which is fine for the 15A Circuit with 14/2.

Now I dim the all 3 dimmers to 50%. Since it is a Rheostat dimmer and not a clipped wave dimmer, the output voltage would be 60V. Using the same formula:

1755W/60V = 29.25A

Why does this not over Amp the circuit? What am I missing?

The only thing I can think of is that the lights are no longer using 1755W.

Thanks

 

[joed]

Wattage is not a constant. That is your error. More accurate constant would be the resistance.
resistance = volts squared/ watts
In your example
120x120/1775 = 12.25 ohms

When you dim to 60 volts.

current = volts / ohms or 60/ 12.25 = 4.9 amps
watts = volts squared / resistance = 60 x 60 / 12.25 = 294
This will be close but not accurate since the resistance will change slightly as the lamps heat up.

 

[Big_John]

Slight correction, it would be ~8 ohms instead of 12.

 

[emtnut]

Wattage is not a constant. That is your error. More accurate constant would be the resistance.
resistance = volts squared/ watts
In your example
120x120/1775 = 12.25 ohms

When you dim to 60 volts.

current = volts / ohms or 60/ 12.25 = 4.9 amps
watts = volts squared / resistance = 60 x 60 / 12.25 = 294
This will be close but not accurate since the resistance will change slightly as the lamps heat up.

I=P/V still works ....
P=1775/2 (@60Volts) = 887.5 watts
so 887.5/120 = 7.4 Amps

 

[Daniel Holzman]

A rheostat is a variable resistance resistor. A rheostat does NOT change voltage, it simply introduces additional resistance into a circuit in series with the loads. So for example, if you had one light with a resistance of 40 ohms, you would draw 120/40 = 3A, and your total power draw would be V x I = 360 watts. If you set your rheostat to 20 ohms, the total resistance of the circuit would be 60 ohms, you would then draw 2 amps through the circuit.

In this hypothetical example, the voltage loss through the rheostat would be V = I x R = 2 x 20 = 40V, and the voltage loss through the light would be V = I x R = 2 x 40 = 80V. The power of the light would be V x I = 80 x 2 = 160 watts, while the power loss through the rheostat would be 40 x 2 = 80 watts, so the total power delivered through the circuit is 240 watts, of which 1/3 is wasted through the rheostat.

Due to the wasting of power by rheostats, modern dimmers use a Triac, which varies the voltage of the circuit rather than the resistance. But the OP said he had a rheostat. The case mentioned by the OP is a little more complex than the example I gave, since the OPS said he had three lights wired in parallel, so the circuit using a rheostat would require analysis of a rheostat in series with three lights, each in parallel. Harder math, I will leave it to the OPS to do the arithmetic.

 

[Big_John]

Unless he's living in 1906, he ain't controlling 2kW of lights with a rheostat.

 

[ER2]

Slight correction, it would be ~8 ohms instead of 12.

Is not that circuit overloaded? Shouldn't it be #12? Don't know US code though.

 

[Big_John]

Is not that circuit overloaded? Shouldn't it be #12? Don't know US code though.

I wasn't referring to the wire size, just an arithmetic error where it should have been 8Ω not 12Ω.

 

[mpoulton]

Unless he's living in 1906, he ain't controlling 2kW of lights with a rheostat.

Yeah, where did this guy find a 600W rheostat? What does it look like and when was it made? That's pretty improbable...

 

[carmusic]

it is a standard old dimmer with triac insides, he probably call it rheostat because it has a round button to adjust light.

 

[ER2]

I wasn't referring to the wire size, just an arithmetic error where it should have been 8Ω not 12Ω.

I agree, but I was just curious about the US code. Is it not overloaded?