I have an electronic heater timer that is operating a relay that feeds a radiant heater. The timer needs at least .5 watts of load to function properly which we don't have with the relay so my thought is to add a resistor to give us the necessary load. The circuit is 277 volt and I would like to know how to calculate this problem.

active| newest | oldestWatts = current X volts or current = watts / volts = 0.5/277= .0018 amps.

Resistance = Volts / current = 277 / .0018 = 153,888 ohms about 150K

Resistor Power = V^2 / R = 277*277 / 150,000 = o.512 W

so use a 1 Watt 150K resistor or two 1/2W 300K resistors in parallel..

The formula for power dissipated by a resistor is just,

P = V^2/R = V*V/R

where V is RMS AC voltage, or time-averaged DC voltage, in volts,

and R is the size of the resistor in ohms.

Your quote, the "circuit is 277 volt", kind of has me scratching my head, because that's not a number I have seen for mains power voltage, on this planet... according to Wikipedia's article on "Mains electricity by country"

https://en.wikipedia.org/wiki/Mains_electricity_by...

But, just for fun, I can do the math with that number anyway. Keep in mind, that if that number, 277 volts, is wrong, then calculations based on it, will be wrong too.

(You know, if turns out the actual number is 227 volts, then you'll want to go back and do the same math, but with V=227 volts.)

Given P=1.0 watt and V = 277 volts, I can solve for R.

R = V^2/P = 277*277/1.0 = 76729 ~= 77000 = 77 Kohm

But the number does not have be exact. A 100 Kohm, or a 50 Kohm, will be easier to find. (In fact two 100K resistors in parallel

isa 50K resistor.) So I am going to use the same formula again, but this time starting with R and V, and solving for P.P = V^2/R = (277*277)/(100000) = 0.76 watt [for R= 100 Kohm]

P = V^2/R = (277*277)/(100000) = 1.53 watt [for R= 50 Kohm]

Some lighting circuits use 277 VAC

Yeah. I guess they do. 277 VAC is the line to neutral voltage in a 480 VAC three phase service, where 480 VAC is the voltage between any two of the three line voltages. Also 277 is 480 divided by the square root of 3.

So 277 VAC power, and 277 VAC loads, are a thing.

Thanks for letting me know this.

why not put a low wattage bulb in parralel with the relay coil.