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Ampere-hour problem

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A 5 volt camcorder now holds 20% fewer electrons than it did two years ago. Today, with a fully charged battery, a user can get 30 minutes of operation before the battery is dead. What was the original AH rating of the battery? The camcorder consumes 1A when operating.

    2. The attempt at a solution
    I'm in my second class of EE and my professor didn't really teach us anything about problems like this. It says the answer is 0.625AH. I started out thinking since it uses 1A for 30 minutes, the current battery AH rating would be 0.5 so the original battery should be 20% more which is 0.6AH. I also did 1A = 1C/s so 30 minutes would be 1800C x 0.20 = 360C + 1800C = 2160C/3600C = 0.6 AH. Am i doing something wrong?
     
    Last edited: Jan 15, 2012
  2. jcsd
  3. Jan 15, 2012 #2
    1A for 30 mins is 0.5Ahr
    If this is what you get with only 80% of the original charge what would 100% equate to?
     
  4. Jan 15, 2012 #3
    Haha wow. so i was just doing the percentage wrong. I set up proportions for a bunch of my other problems but didn't for this one. 0.5Ah/80 = x/100, x = 0.625Ah. Thanks.
     
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