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Can't get this thought out of my head.

Geo-TH,In

Geo-TH,In

Well-known Member
I unpluged my cell phone when the reoccuring thought came back about unpluging transformer adapters when not in use. Transformers are inductive loads, with some resistance. The current in a transformer is out of phase with the current an electric meter can measure. The transformer current is more than the current the meter can measure.

So, if all the above is true, then the heat created inside the transformer, I squared R, will be more than what I'm paying for?

Or another way of looking at it, I would have to use more current to generate the same amount heat if I'm using just a resistor. Therefore, it would cost me more to make heat.

The reoccuring thought is, If I had a large inductive load, like a large transformer connected to a power source, not removing power from secondary, would I be making more heat inside the transforemer without the electric meter seeing all the current in the inductive load?

George
 
I"ve thought about this also; If you have a lamp,
(light bulb) plugged in....but the switch is off,
it"s not using any power. But, if a transformer
is "Plugged in", though there"s no load on the
secondary, there is still electricity "Flowing"
through the "Primary" coil! so how much power do
the small transformers use each month...how much
heat do they put out ?
 
Not sure if this helps any. One time I hooked a Kill-A-Watt up to phone charger. With phone hooked to charger and charging it read 60 watts, when phone reached full charged state it read 60 watts, unplug phone (but not charger) it read 5 watts. Not a huge fantom load but add # of hours each year charger is not in use it adds up. This was a OLD phone and haven't tried same test with a new smart phone.
Amount of heat produced I couldn't feel at all.
 
The issue isn't how much power it's using. The issue is how much of the power is it converting to heat? And how much of the power am I really paying for?

If I got it right, the electric meter measures amps = cosine of the phase angle times the amps in the transformer. Without the transformer being power factored, the cosine of the angle will be less than one.
 
There are two internal sources for heat in a transformer with only the primary connected. They are the actual resistance of the wire coils that form a complete circuit. (no way to avoid that ever) and the second is un resolved Eddy Currents in the laminations and mounting components. These are parasitic voltages created in the iron (or ferrite) of the core and brackets. Really good engineering and manufacturing can limit these but not entirely remove them. Every eddy current is an example of induction heating. Every eddy current also acts as a mini transformer that is shorted in the secondary. This produces in effect an inductive load on the primary, causing incremental phase lag. (probably not measurable without radical lab grade equipment)
So the villan is actual resistance in the winding, but remember in these cold days it is adding to your house heat, and not much of that. Jim
 
I have an old GE AC Milliamp meter used to check the output of low amperage chargers, so I decided to run a test by connecting chargers to the meter to test current draw without load.

18 Volt Drill Charger--------32 MA
14 Volt Power Converter------43 MA
9 Volt Power Conveter-------38 MA

The 14 volt and 9 volt are for a tape recorder or radio of some sort. The meter scale is 0 to 100 MA

So, I can see that some electricity is being used.
 
Finally someone sees my point. How much of the current that you see with your GE AC Milliamp meter will your electric meter see?

It would be nice of you could measure the Powerfactor while the chargers are plugged in.
 
I would say that the resistive load in ma of those chargers is 99.9% of the sum current flowing in the device. The resistance and Phase Lag, power factor, would be way too small to measure, as there is near zero reactive load when the secondary is open. Jim
 
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