I would like to wire an electric baseboard heater and 4 can lights on the same 15 amp circuit. The total load would be about 140 watts under the 80% allowed for the circuit. The heater, which will require 1000 watts, requires 240v hookup. So my question is this, can I safely spice into the 240 line to power the 120v can lights on the same circuit or do I need to run a separate circuit for the lights and a separate circuit for the heater? Obviously, the white and black in the 240v line will be hot, so, if it is permissible to splice into one of the hot legs of the 240v line to power the lights, I don't know what I would do with the neutral going to the switch for the lights.

You cannot tap off a 240 for 120 volts, also the hot needs to go to switch, not the neutral.

You cannot tap off a 240 for 120 volts, also the hot needs to go to switch, not the neutral.


Thanks Strat, that's what I needed to know.

4 - 60 watt bulbs = 240 watts, about 2 amps. The Heater draws 4.166 Amps.
1000/240=4.16666

4 - 60 watt bulbs = 240 watts, about 2 amps. the Heater draws 4.166 Amps.
1000/240=4.16666

I need to figure out the smallest double pole breaker I can use for a 240v circuit powering 2 separate 1000 watt electric baseboard heaters, while still staying within the parameters of the 80% rule. I believe the answer is that I will need a double pole 15 amp breaker. 10 amps would do it, but would run afoul of the 80% rule. Am I correct on this? If so, I will have some left over space on this circuit. Can I run it to an exterior 120/240 combo receptacle?

The current draw of 2000 watts at 240 volts = 8.3 amps x 125% = 10.41 amps is minimum size circuit rating.

The breaker may be the next standard size, which is a 15 amp breaker.

The current draw of 2000 watts at 240 volts = 8.3 amps x 125% = 10.41 amps is minimum size circuit rating.

The breaker may be the next standard size, which is a 15 amp breaker.

Thanks TK. One follow up, I don't understand why you multiplied 125% by 8.3 amps. In any event, my plan is to go with a 15 amp double pole breaker. Thanks again. Ryan

Thanks TK. One follow up, I don't understand why you multiplied 125% by 8.3 amps. In any event, my plan is to go with a 15 amp double pole breaker. Thanks again. Ryan

I think I figured out the answer to my own question. 8.3 amps must be 80% or less than the total load carried by the circuit. To determine the minimum number of amps needed for 8.3 amps to constitute 80% of the total load, 8.3 amps is multiplied by 125% = 10.41. In other words, 80% of 10.41 amps is 8.3 amps. Can someone confirm if this is correct?

The 80% rule is actually for calculating the max load that should be connected to an overcurrent protection device,a circuit rating is calculated by increasing the load by 25% to size the circuit, which is what I did for your heat load.

The 80 % rule only applies to breakers, fuses, and switchgear for certain loads, definitely for electric heating circuits, and load that will run 3 hours or more, which is considered continuous.

Certain loads that do not run continuous can load a OCPD 100%.

The old rule applied 80% to all OCPD for any load. That has been changed a few years ago.

Many times, an older electrician like myself will still apply 80% to all OCPD, as with someone like myself only does commercial and industrial, where most everything runs continuously, so we just apply 80% to all breakers, fuses, and switchgear.

It is an older rule that is embedded in older tradesmen. Just nowadays, to be accurate and factual with younger electricians, and DIY'ers, best to do both calcs.

The 125% rating to size a circuit is still the same as it ever was. Sometimes it works both ways, sometimes it does not work for both circuit ratings and OCPD, so best to do both.