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TWL: Inverters and Volt-Amps questions

From: pw (no email)
Date: Tue Jul 01 2003 - 12:52:07 EDT

  • Next message: Jim McCorison: "TWL: Re: Equipment reliability - Mil Spec"

    Frank Wrote:

    Is there any way to find out how many 12 volt amps it takes to make 1.5
    amps
    120 volts tru the inverter.
    I have a small fridge that is rated at 1.5 ac 120 volts, seems that it draws
    a
    lot from the battery tru the inverter.
    Anyone with similar experience?

    Frank Marinic
    F/V Mar Boa

    Paul's reply:

    "Volt Amps" are Watts. A given piece of equipment will always use [nearly]
    the same number of watts (Volt Amps) whether it's running on 12 Volt DC or
    120 Volt AC.

    A quick way to keep the formula straight in your head is to draw a triangle
    with a "T" in it touching the 3 sides - I'll try to draw it here with text
    characters and hope the e-mail transfer doesn't screw it up.
    (Place the 'W' above the T crossbar - place the 'V' to the left and the 'A'
    to the right of the upright)

        / \
       / W \
      /_____\
     / V | A \
    /_________\

    Remember that triangle and you will always be able to quickly work out any
    situation, solving for the value you need:

    Watts = Volts X Amps (they are beside each other)

    Volts = Watts / Amps (Watts is over Amps)

    Amps = Watts / Volts (Watts is over Volts)

    You will need to allow for losses in your inverter - (I'd guess an allowance
    of 10 - 20%) maybe Arild could give us a ballpark figure there.

    So, in your example of 1.5 amps and 120 Volts - the power usage is 120 X 1.5
    (Volts X Amps) = 180 Watts.

    Now change the Voltage to 12 volts on the battery side of the inverter. Use
    the formula again. The watts haven't changed - the new voltage is 12 - so
    solve for the new Amps value.

    The amps would be 180 / 12 (Watts divided by Volts) = 15 Amps. Add 10-20%
    for losses and you have your answer.

    Don't forget that when you are working with batteries, time becomes a factor
    as well, because now you are talking about Ampere-hours. (1 amp flowing for
    1 hour) As currents get higher the relationship to a battery's capacity is
    not linear - i.e. The efficiency goes down - maybe Arild could refresh our
    memory there also - he spoke about this in a post a week or so ago when
    talking about batteries and heavier currents - I think it was "Peukert's
    Phenomenon" [nothing to do with seasickness])

    Cc: Arild, Frank

    -- 
    Regards, 
    paul whealy
    _________
    The trouble with the rat race is that
    even if you win, you're still a rat.
                            -- Lily Thomlin 
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  • Next message: Jim McCorison: "TWL: Re: Equipment reliability - Mil Spec"



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