Stopwatching your meter
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7:23
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most people don't what an ammeter is. That is true.
meterdatamanagement 7 months ago
Like I said over a year ago though, it's not very practical for any homeowner or electrician to do this. So the question is basically moot.
gyneman 2 years ago
your ammeter is basically a current transformer showing amperage for each leg. clamp each leg individually, / your voltage for each leg and add both wattages together, giving your total wattage at time of metering. if you know your Kh premium, you can figure out your utility cost at that very moment of metering. As your usage varies so will your amperage. Making it vary difficult to track. Which is why your power company uses a wattmeter.
gyneman 2 years ago
Not every home owner has an ammeter and is comfortable clamping an ammeter around a service conductor.
In addition, the homeowner would have to know when to clamp both hot legs of a 240 V service and divide the sum of the amperage measured in each leg by 2 to get the overall amperage load or to just clamp the hot leg of a 120 Volt load.
They would have to know that I X E = P to get the apparent power.
Clamping would not be accurate as a bill is based on true power and not apparent power
hdfatboy2k 2 years ago
KW LOAD= (Kh)(Revs )3600 / 1000(Time)
As simple as that guys. Your load = Kh times amount of revs counted times 3600seconds (sec in an hour) all devided by the quotient of 1000(Kilowatts) times the time recorded to get the counted revs.
This changes if you have CT (current transformers) and a DK (Dial constant) other than 1.
The DK is a multiplier for the dials. If you have a DK of 2 then the register is showing 1/2 the actual usage and must be multiplied by 2.
camsixseven 2 years ago
/watch?v=GoT2Epl7sGM
hardtoremember 3 years ago
that was so fresh dude thanks a grip
DealerDirekt 3 years ago
Hi..
Is your method and does your "calculator" work for other countries as well? I am from the Philippines..
TheBrownChiLD 3 years ago
your website has expired!
tigerbody69 3 years ago
hey meterguy! i bought a device that reduced my power by 40% want to know more ASK me
MeterBeater 3 years ago