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I was just now running trains and thought I would check volt/amp numbers out of curiosity. I got different numbers at the same time while running my old SF F3.  Would someone explain why?  Is this difference at all significant?  Thx for your time. 

VOLT/AMP

AMPS

TRACK VOLTS

 

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  • VOLT/AMP
  • AMPS
  • TRACK VOLTS
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I thought I had an answer but....  just to be clear: ZW-L analog meters read 15 volts and 4 amps, digital meters read 0.73 amps (hard to read)  and 11.29 volts???

Are you sure they are reading at the same electrical points? There has been a lot of discussion on the forum on chopped sine waves and how that impacts various meters.  However, Seems hard to imagine chopped sine waves (if that is what the ZW-L produces) can explain 4 amps versus 0.73.

Napkin maths starting with the kill-a-watt meter. This is the current draw on the 110v in your house which was  0.73A. So 110v x 0.73A = roughly 80 watts of draw going into the ZW-L. 

Napkin maths for the ZW-L to the track is 15v x 4A = giving us 60 watts.

That sounds about right considering the ZW-L is using 60 of the 80 watts it's getting from the wall as track power. The remainder of 20 watts is used on the internal circuitry doing all sorts of stuff running the gauges and legacy systems and we lose some power to the step down as heat too.

As for the meter, a normal handyman meter has trouble reading the power output wave of the transformer (or so I've read), I wouldn't worry about the few volt difference the meter is giving you from what the Transformer gauges are showing. I've read that a true RMS meter more accurately displays the voltages coming off of our transformers, but this is just what I've read on the internet so take that with a grain of salt.

So to sum up and answer your original question, the differences you are measuring are not significant but rather just the difference if how we see the power used. Given a fixed load (Wattage draw/demand) as voltage lowers, current rises and the reverse is true as well as voltage raises current lowers, hence why we see 0.73 amps on the 110v outlet, and 4 amps on the 15v track power. The equation we are observing here is as follows:

P = I x V

Where P is the wattage in Watts, I is the current in Amps and V is the Voltage in Volts. From there its all just algebra depending on what we know and what we are trying to find out! 

Last edited by Phoenixx101

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