Resistive Load - Amp calculation question.

Resistive Heating element:

Spec = 240VAC 1200 Watts.

Amps calculates to 5 AMPS.

If the same load was powered by a 208VAC source, AMPS would calculate to 5.7 AMPS.

Question is, is the WATTS rating printed on the label a constant? In other words, regardless of the voltage applied, we should be using MFG's WATTS rating to calculate current draw based on the supply voltage. Trying to understand how "WATTS" remains a constant in this scenario un-affected by V or I.

Cheers.

The calculation for electrical power is P=U*I. This means I=P/U and leads to 1200/240=5A and 1200/208=5,769A.
As you can see - it's not constant. The rating only means that if you have 240V then it needs 5A.
Little bit different with motors but for heating elements relative simple.

Regards
Thomas

P = U^2 / R
R = U^2 / P = 240 ^2 / 1200 = 48 Ohms – heater resistance
I = U / R = 208 / 48 = 4.33 Amps

As the others have sort of said, in a round about sort of way. If your heater is rated at 240V a.c., then you won't get 1200 watts powering it at 208V a.c., you would get 1040 watts.
And there's more . As the heater gets hotter its resistance will change, generally the resistance rises, so your watts will reduce again. It depends on the heatimg element temperature coefficient on how big the change will be.

Wow go back to basics
The rating are 1200 watts with a 240 volt supply
The constant here is the resistance of the heater
The amount of current would depend on the supply voltage
Ohms Law is what applies here
The constant here is the resistance of the heater
Resistive load
240 volts 1200 watts = 5 amps 1200 / 240 = 5 5 amps
240 / 5 = 48 48 ohms
208 / 48 = 4.333 4.333 Amps
At 204 volts the heat produced would be 901.3 watts
480 volt supply
480 / 48 = 10 10 amps
At 480 volts the heat produced would be 4800 watts
You also have to assume that this is a pure resistance load and you have no inductance to consider
You also must keep in mind that assuming that the 240 – 208 volt source is AC then the 1200 watts would be the average watts because the supply voltage is 240 V RMS

That has me scratching my head , Friday brain fog.
The reduction in wattage should just be the factor of new volts divided by the old volts, simple ratio.
But GaryS maths is spot on, and gets a different result.
I will be worrying about this all day

BryanG said:

That has me scratching my head , Friday brain fog.
The reduction in wattage should just be the factor of new volts divided by the old volts, simple ratio.
But GaryS maths is spot on, and gets a different result.
I will be worrying about this all day

Click to expand...

W is a function of V and A, not just V. Reducing V also reduces A, given constant R.

Or, using W = V * I and V = I * R
I = V/R
W = V * (V/R)
W = (V^2)/R

Also:
W = (I*R) * I
W = (I^2) * R

So, reducing the voltage reduces the power by the square, not just the straight ratio of the reduction.

BryanG said:

The reduction in wattage should just be the factor of new volts divided by the old volts, simple ratio.
But GaryS maths is spot on, and gets a different result.

Click to expand...

How wattage will change depends on what is being held constant.

Since R is constant and we are varying V, use Watt's law formulated as W=(V^2)/R (we could also use W=(I^2)*R, but then we'd have an extra step calculating the current for no benefit).

The ratio between the old and the new is the ratio of the squares of the voltages. Use this and your math will agree with GaryS's.

EDIT: Beaten to it

Pretty standard theory that most junior electricians learn. If you wire up a 240V resistive heater incorrectly with 120v, you only get 1/4 of the power due to the formula other's have mentioned:


Power = V^2/R


This is different than an inductive load such as a motor that will compensate for lower voltage by pulling more current. Resistive loads work differently.


So yes, if you powered this 240v heater with 208V instead, you'd be pulling 4.333A at around 901 watts.

Happy that this old man's Friday afternoon brain fog brought you such entertainment
You just wait till I get my perpetual motion engine prototype working