Rank: Advanced Member
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Joined: 9/1/2008(UTC)
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Again ! Where are you taking your readings. You post point nine.
This unit is rated 7.2 amps It is impossible if the unit were not properly charged to have the same amperage draw as when properly charged.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
Posts: 440
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Rank: Member
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Joined: 10/26/2008(UTC)
Posts: 24
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I may not know anywhere near what you know about refrigeration, but if there is one area of expertise that I have, it is that of electricity and/or its related physics.. I have been taking ammeter, voltage, resistance, capacitance, etc. readings for as long as you have been dabbling with refrigerators.
I indicated in one of my prior postings that I had taken other readings with same meter to make sure I was doing it correctly - my methods and readings were consistently correct with the other gear that I tested.
If the unit is rated 7.2 amps I would not be surprised if that specification is a worst case scenario, i.e., when the unit's other peripherals are fully engaged. For example, this morning I was surprised to see that the meter was pushing three amps from out of no where! Then I heard the dropping of ice cubes and the onrush of water and the meter went back to the normal .9 reading. The jump was explained by the heater in the ice maker!!!!!!!!
And when I remarked to the repairman about the low consumption he remarked that many new refrigerators units pulled amperage along the perameters I cited. My Sears Kenmore pulled a constant 6.8 amps but that was about 20 years ago.
Thanks much for all of your help and assistance; it has been deeply appreciated. The refrigerator is working fine now. But the interesting question remains unanswered. Maybe somebody else out there hs a clue.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
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The total running amperage of this Refrigerator is 7.2 Amps.
The compressor rating is 6.875 Amps when fully charged, so where you are taking your readings of .9 amps is out to lunch.
I am also a "Licensed Electrician" in the county of Suffolk, so I have just a small clue of what I am talking about.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
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To determine the Wattage (power) multiply ExI, whereas E=Voltage and I=Amperage then the sum is W or Wattage.
If you measured 123.7 E and multiply .9 I (Your measurement) W is 111.33
By Maytags Specs, 123.7 E times 7.2 I equals 890.64 W
According to Maytags Specs, the compressor alone is rated at 825 Watts, so someone should swallow their pride and admit they are wrong.
E 123.7 that you measured divided into 825 W determined by Maytag and Embracco (compressor manufacturer) equals 6.875 I . So a reading of .9 is along way off.
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Rank: Member
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Joined: 10/26/2008(UTC)
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I am fully familiar with Watts law of I = V/R and there is no need for you to lecture me on it no more than I should lecture you on the concept of back EMF in electric motors.
The meter that I use has a special adapter that you plug appliances into and you get the reading without having to drape the meter around one leg of the wire going to the applaince or motor.
How do you explain the fact that when I plug a toaster oven into it I get a reading of approx 13 amps, and an iron goes for about ten amps and my old Kenwood gave me a reading of almost 7 amps. And a 100 watt bulb goes for about an amp. These are all readings using the same meter.
Now please answer the question directly - i.e. - how can you explain the
fact that the meter gives me correct readings for these appliances and not the refrigerator when they all get hooked up in the exact, same, identical way !?!?!?!?!? EXPLAIN, if you can.
Meanwhile if unit was pulling .8 KWH, since unit never turned off, it was using 19.2 KWH daily or about 576 KWH for a 30 day month. At .14 per kwh down here we are talking about a contribution of $ 78.14 towards my monthly bill of 190 bucks. The numbers simply do not add up nor support your contention.
And it is not a matter of pride. I freely admit to not being a know it all.
That is why I humbly ask questions and always look forward to learning more. I simply thought that you knew lots more about the matter than I did and I asked. And I was wrong and I am sorry that I did ask and put you in an uncomfortable position.
I will find out from someone else and when I do, I will be delighted to share the knowledge with you. Ok? No big deal.
Just chill, and thanks again for your assistance.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
Posts: 440
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First I = E/R is OHMS LAW not WATTS.
And yes the device you are using is called a "line splitter" which I also use.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
Posts: 440
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  The ALS1 Line Splitter for clamp-on meters eliminates the need to split electrical cords or open electrical boxes for current measurements on 120V lines.Features • 1 to 1 and 10 to 1 internal coils for better resolution • 15A max • Voltmeter input jacks • Integral ground conductors • One year limited warranty
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Rank: Member
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Joined: 10/26/2008(UTC)
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I realized that I referred to ohms law by error after I sent you transmission. I knew that watts equals watts times volts. My boo boo.
Anyway, good to hear that you also use a line splitter like I do. But you still have not answered the question I posed, and have chose to ignore it completely. All U did was tell me that I cited the wrong law. Bully.
But after supporting my question with all of my measurements and observations do you have any valid explanation for why my readings with a fully charged system are still .9 amps. And if you do not know the answer or have an explanation why not just admit it rather than just ignore and nitpick? I gave you all of the statistical data you need and assured you that I was aking proper readinbgs.
What is wrong with not knowing everything? I certainly do not, and that is why I ask questions. Do you have an answer for the last time because the situation just does not make sense. And if the drier was clooged, then it would have had more pressure and now that I have a new unclogged one the pressure sure be less along with less poweruse.
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Rank: Advanced Member
Groups: Senior Expert
Joined: 9/1/2008(UTC)
Posts: 440
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You cannot I repeat not have an "Amperage" draw of point nine from a refrigerator that has a compressor rating alone of six point eight seven five amps. With a total rating of seven point two.I do not claim to know everything, that is why I back up everything I stated with manufacturers technical literature.
So if this is the case, how are you calculating KWH ? (Power Consumption)
A restricted drier will feel "cold" to the touch indicating the pooling of refrigerant.
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