Pulse motor - is this overunity?

[Groundloop]

@hartiberlin,

You are right. I lloked at the circuit drawing. The drawing says 330.

EDIT : Very poor drawing but I think it says 33 Ohm. 

So just divide the output by ten = 0,09026457 Watt. 

Groundloop.

[mr2]

Tanks for answers.

First of all:
If R2 "sees" the resistance R3 at 33 ohms, why don't the R1 see it?
R1 is dependent of the R3 as much as R2.

Take away the "black box" and explain it then out of the same values.


I have the same voltage out. I have higher voltage over the shunt-resistors at the output than the input. That says: More amperage at same voltage at the output.
No overunity?

[mr2]

@hartiberlin,

You are right. I lloked at the circuit drawing. The drawing says 330.

EDIT : Very poor drawing but I think it says 33 Ohm. 

So just divide the output by ten = 0,09026457 Watt. 

Groundloop.

The drawing and my initial message says 33 ohm.

But the R1 is as much dependent of the R3 as R2. Don't know why you both calculate with the 33 ohm resistor, when the shuntresistors measure power on the "same line".

EDIT: btw: the formula says P=U^2/R. Then I have it correct in my first message. (Well.. i did know)

[hartiberlin]

@Mr2,
in your first posting you did only
compare the wattage at the shunt resistors but not the
input power from the battery versus the output power at
the 33 Ohm load resistor.
These are 2 very different cases.

[mr2]

@hartiberlin

I understand that. But taking the voltage at both battery and output in consideration, what then?

If my initial drawing is not correct measurement, i could use ordinary amperemeters instead. Easier then. And put a voltmeter at the battery, and at the output.

But out of the shunt voltages groundloop calculated the amperage. I use them in the P = U * I formula.
The measured voltage on battery is 11,62 V, and at output is it 11,6 V.

Then I have:

At R1: 0,0367 A * 11,62 V = 0,4264 Watt
At R2: 0,0523 A * 11,6 V = 0,6067 Watt

Hmm.. THIS must be correctly measured and calculated.