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Q on battery voltage

  • 29-06-2005 7:06pm
    #1
    Bri
    Closed Accounts Posts: 2,035 ✭✭✭


    Thanks to your help in here I've finally got a reply from studentcomputers.co.uk. They tell me that although my particular battery number (289053-001) for my Presario 900 isn't listed, that http://www.studentcomputers.co.uk/laptop-battery/compaq-laptop-batteries/compaq-evo-n160-presario1700-17xl-series.htm is the right model.

    Can anyone advise me independently if there will be a problem as:
    My old battery says 14.4V and 4.4AHr
    The new battery says 14.8 Volts, 4000 mAh?

    I'm outa my depth when it comes to this kind of stuff.

    Thanks in advance.


Comments

  • #2
    ARGINITE
    Registered Users, Registered Users 2 Posts: 2,718 ✭✭✭ARGINITE


    It states on the website that it is compatible with Compaq Presario 900, so bob's your uncle.
    As for the little difference in voltage and mAh. That shouldnt matter at all.


  • #3
    Bri
    Closed Accounts Posts: 2,035 ✭✭✭Bri


    D'oh. Sorry my mistake. I know what I did - I just searched immediatley for the matching part number.

    Why does it not matter? Is it the fact the numbers are very close or something different? (interested to know).

    Thanks!


  • #4
    ARGINITE
    Registered Users, Registered Users 2 Posts: 2,718 ✭✭✭ARGINITE


    Well the "4000 mAh" has to do with the max amount of current that can be drawn from the battery for one hour. So the lower number just means that it can draw 4400mA from the old battery for one hour and 400mA less with the new one.
    That means that the older battery would last a little longer than the new one.

    As for the 14.4V for the old battery and 14.8V for the new battery.
    This shouldn't matter as it's a little higher than the old battery voltage and their should be voltage supply circuit in the laptop, that would convert this voltage down to the required voltage of +5V,+12V,-12V,+3.3V.

    Why the voltage being a little higher.
    Because if it was a little lower than the battery would not be able to supply enough voltage and therefore with Ohm's law *(R=V/I) the lower voltage(V) would be made up for with a decrease in current(I).

    *Resistance = Voltage / Current. (Ohm's Law)

    Hope that answers your questions.


  • #5
    Bri
    Closed Accounts Posts: 2,035 ✭✭✭Bri


    That does indeed; even reminded me of a little L.C. physics while you were at it!

    Much obliged, thanks to you I can quit hogging all the seats near plugs when I work :D Seriously though - thanks, this should make a big difference to my work.

    +rep


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