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Calculating Charge (Q)

  1. Mar 23, 2010 #1

    FeDeX_LaTeX

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    Hello;

    I just wanted to clarify whether I am using this formula correctly. By definition, current is the rate of flow of charge. In other words, its how much charge flows per second. One amp (1 A) is equal to one coulomb per second (1 C/s). Charge and current are related by the equation: I = ΔQ/ΔT.

    1) A battery supplies 10 C over a period of 50 seconds. What is the current?

    My answer: Given the formula above, I = 10 coulombs / 50 seconds = 0.2 A. Is this correct?

    2) Another battery is connected for 2 minutes and provided a current of 0.4 A. How much charge flowed?

    My answer;

    0.4 = Q/120
    Q = 0.4*120 = 48 C

    3) A car battery has a capacity of 24 Ah (amp hours). If it provides a current of 48A how long can it be used for? How much charge (in coulombs) does it contain?

    My answer: Not completely sure about this question, but I think that the car battery can be used for 2 hours. As for the latter question, I don't understand... how do I convert from amp hours to amps?

    Thanks.
     
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  3. Mar 23, 2010 #2

    kuruman

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    Your answers to 1 and 2 are correct. An amp-hour is current multiplied by time. What physical quantity do you get when you do that?
     
  4. Mar 23, 2010 #3

    FeDeX_LaTeX

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    Hello;

    Yeah I thought about that but I ended up getting 1 amp per 150 seconds, which isn't much help... but 1 amp is 1 C/s. How can I use that in my formula?

    Also, I said that the battery can be used for 2 hours, but is it actually 0.5 hours? If the battery has a capacity of 24 Ah, so if 48*time = 24, 24/48 = 0.5?
     
  5. Mar 23, 2010 #4

    kuruman

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    Once more, what physical quantity do you get when you multiply amps with hours?
     
  6. Mar 23, 2010 #5

    FeDeX_LaTeX

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    Oh... 1 coulomb. How could I not see that...

    So is the answer 2 coulombs?
     
  7. Mar 23, 2010 #6

    kuruman

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    No. To get coulombs, you need to multiply amps by seconds, not hours.
     
  8. Mar 23, 2010 #7

    FeDeX_LaTeX

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    2 hours = 3600*2 = 7200 seconds.

    So the answer is 7200 coulombs?
     
  9. Mar 23, 2010 #8

    kuruman

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    Why are you multiplying by 2? What is the rating of the battery in amp-hrs?
     
  10. Mar 23, 2010 #9

    FeDeX_LaTeX

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    Whoops. If it provides 48A, then it will be using twice as much current and thus will last for half the time, so 0.5 hours. 0.5 hours in seconds is 1800 seconds, so is it 1800 coulombs?
     
  11. Mar 23, 2010 #10

    kuruman

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    One more time. To get Coulombs you multiply amps with seconds. You have 1800 s, how many amps does the battery provide during this time?
     
  12. Mar 23, 2010 #11

    FeDeX_LaTeX

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    48*1800 = 86400 coulombs
     
  13. Mar 23, 2010 #12

    kuruman

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  14. Mar 23, 2010 #13

    FeDeX_LaTeX

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    Little bit unrelated, but Watts work in the same way, correct? Watts = Joules / seconds, so to find J I'd have to multiply watts by seconds...?
     
  15. Mar 23, 2010 #14

    kuruman

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    Precisely.
     
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