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How long will my battery last?

  1. Aug 12, 2011 #1
    Hey guys!

    I have two camcorders.

    Cam #1: Power Supply: DC 7.2V (battery)
    Power Consumption: 7.4W

    Cam #2: Power Supply: DC 7.4V (battery)
    Power Consumption: Doesn't say :( - but should be the same roughly.

    How long can a "portable powerpack" with the following specs keep these camcorders running before running out of power?:

    Capacity: 1000VA / 500W
    Nominal Voltage: 120V / 230V
    Battery Capacity: 12V / 9Ah x 1pc (think it says 1pc cause it's a UPC for computers?)

    Could someone please let me know the math on how to figure this out myself for future equations, too?

    Thanks so much!
  2. jcsd
  3. Aug 12, 2011 #2

    The potential difference/voltage between the battery poles tells you how much energy (in Joules) is available per unit of charge (1 As). Short: Higher voltage means a single charge carrier can do more work for you.

    A charge capacity of 9 Ah tells you, that the battery can drive a current of 9 A for an hour, or:
    that it stores 9 A * 3600 s = 32400 As of charge. A unit of charge can do 12 J of work (12 V = 12 J/As) and you have 32400 units of charge. Consequently you have 12 V * 32400 As = 388800 J available.

    Your camera consumes 7,4 W = 7,4 J/s on average. Which tells you that it should run for 388800 J * s / 7,4 J =52541 s = 14,6 h. Which is a lot, but then again you have a very big battery - capacity is on par with a car battery.

    I do not understand these two lines though:
    I am guessing that your 12V battery has a built in 12 V DC to 120 V or (switch) 230 V AC converter - since you said it is a PC reserve unit. So the second line represents possible output voltages and the first one maximum power output respectively (1000 W at 120 V and 500 W at 230 V). Be sure to use an adapter/converter (120 V to 7,4 V or 230 V to 7,4 V) or you will roast your equipment.

    It would also be better (more efficient) to bypass the built-in converter and go straight from 12 V to 7,4 V, because converters are not perfect and each one degrades battery efficiency (battery time).
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