Gerard Morin's Replication with Eleman Magnet Motor Test 1

[ELEMAN]

what can you say about this picture ? The first picture has effects on camera....BUT LOOK THE VIDEO www.youtube.com/watch?v=jAcbZ6CvDTc&list=PLT7Zi_DRtP7fpdEHDlFIfaI_bOwZo5zyF&index=21

[MarkE]

I can say the white levels are different in each.  In each case the fluorescent bulb appears much dimmer to me than I would expect from ~10Watts of input power.

Personally, I dislike the use of light bulbs of any kind as a proxy for power.  But if you are intent on using a light bulb in front of a camera to try and indicate power, then I suggest that you run two bulbs side by side where one is a reference driven by a power supply adjusted such that the light output of both lamps is as close to equal as possible.  The much better alternative is to perform proper power measurements of both the input and the output.

[Groundloop]

Your measuring the output voltage on the wrong side of the transformer.

You Ampere meter is on the input side of the transformer and
your Volt meter must also be on that side. So the transformer
is a 230 / 9 ratio = 25,6. 125 Volt / 25,6 = 4,9 Volt. So the
output = 4,9 * 200mA = 0,98 Watt. Input = 24 * 0,38 = 9,1 Watt.
COP = 0,98 / 9,1 = 0,1.

GL.

[MarkE]

I agree that the voltage and current should be measured on the same side of the transformer, preferably the output side.  I agree that the transformer turns ratio means that the output current is no more than 9/230 * 200mA <= 7.83mA, which at 125V <= 0.98W into the fluorescent bulb, or alternately that the transformer input voltage is: ~9/230 * 125V ~= 4.89V which at 200mA is also 0.98W, under the favorable assumption of a unity power factor.  And I agree that the input power is:  0.38A * 24.0V =  9.1W making the power to the fluorescent tube less than 11% of what is drawn from the power supply.  The less than 1W power accounts for the low tube brightness.  How much less than 1W / 11% the fluorescent tube power is we do not know because we do not have any phase information.