220VAC Power Questions

[Anthony Whitesell]

Power factor of what? The motor is running, so the power factor is either 1 or 0 (don't remember exactly what the equation is). Two meters. Handheld and wall powered, 390 and 380mA respectively.

[Dan Friedrichs]

In practice, power factor is not ever going to be exactly 1 or 0.

Rod is saying that the current you are seeing on your meters may be out-of-phase with the voltage present, meaning that the true power being consumed by the load is less than V*I. The cosine of the angle between the voltage and current phasors is known as the power factor. For a purely resistive load, the power factor will be 1 (or nearly so). For a more inductive load (like a motor), the power factor could be much lower, meaning that the power drawn by the motor is actually much less than V*I.

I'm not sure the problem is really clear at this point. As I understand it:
1) You have a jointer that has a magnetic starter.
2) The magnetic starter has a small light on it to let you know that it's plugged in.
3) When the motor was off, you measured the current through one leg (how did you do this?) of the 220V circuit, and measured appx 0.4A
4) You are concerned because 220V * 0.4 A = 88 Watts, which is quite a lot of power to be wasting 24 hrs a day (I agree).

If all the above is true, I'd be concerned that something was wrong with the starter. When the motor is off, nothing in the starter (except that light) should be drawing power, either. When the motor is on, there will be some small amount of current used to energize the relay coil, but only when the motor is on. I'd consider disconnecting that little lamp and measuring the current again to see how much its contribution is.

[Anthony Whitesell]

Knowing that I was not going to actually run the machine, I pulled back the cord cap, removed the white from the plug and inserted the meter between the plug terminal and the end of the wire.

[Dan Friedrichs]

Well, that's the way to do it...

I'm really surprised - that's a lot of power for an indicator light. Maybe Rod will have another suggestion, but I'd open the switch box, take the light out of the circuit, and repeat the amp measurement. Something doesn't seem right, here...

[Chip Lindley]

Back to the OP's current drain:

Yep. Surprise! I measured one leg of the 200 while the motor was off and it was drawing 400mA. I don't know what is in the control box other than the red light. Unplugging it is getting a little annoying, but atleast it is not a difficult reach.

Understandable IF the magnetic starter is a low voltage-type with a 24V transformer in the control box. That xformer will be powered when the red light is on. Delta's LVC controls used a .040 Kva multi-tap xformer to power the 24v coil.

If your mag starter's coil uses line voltage, nothing is energized and no current should be flowing until you push Start. In this case, there may be a voltage drain somewhere besides that little pilot light.

Ok, so I ran a dedicated 240V 20A circuit under the floor and up to a box mounted on my 3HP Unisaw and DJ-20. I want to install a switch simply for disconnect...changing blades, general safety. I have two Leviton CSB2 DP 15/20A Switches that were listed as handling motor loads. I'm good?

Go whole hawg and use a commercial 30A 2-pole disconnect. They are cheap enough on eBay:

Square D Light Duty Disconnect Switch L211N 30A
SqD 30A disconn.jpg

[George Bregar]

Go whole hawg and use a commercial 30A 2-pole disconnect. They are cheap enough on eBay:

Square D Light Duty Disconnect Switch L211N 30A
SqD 30A disconn.jpg

Thanks Chip...I looked at my original post and it's worded poorly. To clarify, both the Uni and DJ are on their own dedicated 240V 20A circuit if that changes anything. These are 30A...why? Because of the 3HP uni? The DJ-20 is 1-1/2 HP. WHy a fused disconnect? Could I use this http://www.homedepot.com/h_d1/N-5yc1...atalogId=10053