Re: Identifying Current and Voltage on vaguely marked MOT. (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Mon, 27 Aug 2007 12:46:52 -0400
From: David Speck <Dave@xxxxxxxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: Identifying Current and Voltage on vaguely marked MOT. (fwd)

Glen,

Watts is (approximately) volts x amps.  Not entirely straightforward in 
setups where you are driving inductive loads where current and voltage 
are not precisely in phase, but close enough for our approximation. 

If you determine the output voltage of your MOT, often by driving the 
secondary with a known voltage, and then measuring the voltage on the 
input windings, then you can determine the turns ratio and calculate 
what the nominal output voltage will be when the primary is driven by 
120 VAC.  Divide 700 by the voltage, and you will get an estimate of the 
rated current.

A MOT can deliver more than its rated current for a short time, but will 
get hot faster if overloaded, and it may burn out.  Also, MOTs come with 
magnetic shunts which will provide a degree of current limiting, but not 
enough to withstand a prolonged short circuit like a NST can sustain. 

There is also a formula to estimate the wattage capability of a 
laminated steel core transformer based on the cross section area of the 
center of the core.  VA Rating = ((Core Cross section area (sq in))/0.16)^2.

HTH,

Dave

Tesla list wrote:
> ---------- Forwarded message ----------
> Date: Mon, 27 Aug 2007 00:21:46 -0500
> From: Glen McGowan <glen.mcgowan@xxxxxxxxx>
> To: Tesla list <tesla@xxxxxxxxxx>
> Subject: Identifying Current and Voltage on vaguely marked MOT.
>
> Just curious if there's some super secret math out there that can help me
> identify the rated voltage and current of a MOT.  I only have one variable
> for the equations.... the MOT is labeled with 700W output. That's it......