# Need to create a dummy load

1. Jan 14, 2010

### Medalcraft

I need to create a dummy load on a 120VAC circuit. The load must draw .3A. How can I do this?

2. Jan 14, 2010

### mgb_phys

A resistor?

If you don't know Ohm's law to work out the resistance and how to work out the power in the resistor are you sure you should be playing with 120VAC?

We aren't trying to be difficult but if you are an 8year old kid somewhere who has decided to learn electricity with potentially lethal line voltages because they don't have a spare AA battery we don't really want to be responsible

3. Jan 14, 2010

### vk6kro

You could get four 10 watt 100 ohms resistors and put them in series.

Total resistance = 4 * 100 = 400 ohms

Current = 120 volts / 400 ohms = 0.3 amps

Power dissipated in each resistor = 0.3 A * 0.3 A * 100 ohms = 9 watts

Total power = 0.3 A * 0.3 A * 400 ohms = 36 watts

4. Jan 14, 2010

### Medalcraft

Thank you. That is what I thought, 400 ohms. Just wanted to make sure. I tried a different resistor that was here, 1K - magic smoke.

Forgot to factor in wattage rating for resistor.

Last edited: Jan 14, 2010
5. Jan 14, 2010

### mgb_phys

No it wasn't powerfull enough.
A regular little resistor can dump about 1/4 watt of heat - any more than this and it will just melt

A 10W resistor is a larger more expensive lump of metal it wil look something like this

It's also going to need mounting to some sort of heatsink to dump the 40W of heat

6. Jan 14, 2010

### Medalcraft

Thank you again. I forgot about the wattage values for resistors. Been quite a few years since I last dealt with resistors.

7. Jan 14, 2010

### vk6kro

If 0.33 amps would be OK, you could use a 120 volt 40 watt lamp bulb.

Power = voltage * current
so current (in amps) = power (in watts) / voltage

Current = 40 watts / 120 volts = 0.33 amps.

8. Jan 15, 2010

### MATLABdude

vk6kro, I bow to your Alexander-esque knot-chopping abilities. To the OP, this might also be easier to wire up, since you can go down to your local hardware store, buy a light socket, and then use wire nuts to connect it to your circuit (if applicable).

9. Jan 15, 2010

### sophiecentaur

The only snag with using light bulbs is that you can't be certain of their resistance at anything other than their normal operating temperature. The resistance varies over a range of ten to one from hot to cold. You may need to experiment a bit and even use two in series (one high wattage and one low wattage) to get the resistance you want - if its value is critical. You can also buy very low wattage heaters which don't operate at white heat and will have more stable resistance values.

10. Jan 15, 2010

### vk6kro

This is a 120 volt lamp being used on 120 volts. It should be OK unless the slight increase in current (0.33 A vs 0.3 A ) matters.

11. Jan 27, 2010

### mheslep

Well that 'cold' low resistance holds for only ~.1 seconds, after that time rated current and wattage is expected. So yes there will be a ~10X startup current for the OP, then quickly settling to 0.3 A and staying there absent extreme ambient conditions.

12. Jan 28, 2010

### sophiecentaur

My comment really related to the possibility of the supply volts not always being the nominal value - the resistance varies right over the filament temperature range. Medium / High power resistors are not expensive are they? It all depends on the accuracy required, in any case.