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Please Help Me THIS IS DRIVING ME MAD

  1. Nov 18, 2011 #1
    Please Help Me!!! THIS IS DRIVING ME MAD!!!

    V/R=I Simple Ohms law right?

    This is what I don't get I've been searching everywhere & I still can not find the answer!

    If I were to have a 12 volt battery at 5 amps & if I were to apply a 1 ohms resistor to it.

    12v/1ohms = 12 amps???!!!

    So by adding a 1ohm resistor to my battery I magically increased my amps??!

    Please some one help me...
     
  2. jcsd
  3. Nov 18, 2011 #2

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Check your equation: V = I * R
     
    Last edited: Nov 18, 2011
  4. Nov 18, 2011 #3
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    I meant to say v/r=I still would equal 12 amps.
     
  5. Nov 18, 2011 #4
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Battery only specified in voltage. They specify amp hour as the energy storage in the battery. They do not specify "at 5A". You miss read that.
     
  6. Nov 18, 2011 #5
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    No actually my battery states very clearly 2,600 mili amps per hour. Which equals 2.6 amps.
     
  7. Nov 18, 2011 #6
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Another question is doesn't adding a resistor decrease both my volt & amps of my battery? & How can I calculate that?
     
  8. Nov 18, 2011 #7

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    It's just a way of specifying the energy available. It still comes down to amp*hours. You will get somewhat less available energy at high current drain rates, and somewhat more available energy at low drain rates.
     
  9. Nov 18, 2011 #8

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Adding a resistor where? V = I * R, so the output current is determined by the battery voltage and the resistance across the battery.
     
  10. Nov 18, 2011 #9
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    I hooked up 3, 2v 8amp batteries in parallel & when I tested my amperage I got 24 amps at 2volts. I created something that required a very high current at 16 amps & a low voltage at 2volts. The Amperage listed on batteries is very precise & matches what I get in amps per hour.

    For instance when I hooked up a 8 amp 2volt battery to my array of capacitors it took me 30 seconds to fully charge them. When I hooked them up to a 24 amp 2volt battery I was able to charge my capacitors in 2 seconds.

    Thus the more amps provided per hour the larger the capacity of a battery.
    I am sure you already know this but still does not answer any of my questions.
     
  11. Nov 18, 2011 #10

    cmb

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    :eek:

    So... err.!! If I drive at 26 miles an hour away from my house, then I must be 26 miles from my house?
     
  12. Nov 18, 2011 #11
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    If I were to add a 1ohm resistor on the positive line/wire that is hooked up to a battery, then test for amps & voltage.
     
  13. Nov 18, 2011 #12

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Sorry, none of your math works, and it doesn't match how batteries and capacitors work.

    You said your 2V "8A" battery charged the caps in 30 seconds, and your 2V "24A" battery charged them in 2 seconds. 24/8 does not equal 30/2.

    Please look at the datasheet for your battery (you can usually find it online at the battery manufacturer's website). It will explain what the Amp*Hour capacity means, and how it varies with discharge rate.

    Then always use V = I*R in your calculations. With capacitor charging, the equation is different.
     
  14. Nov 18, 2011 #13

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    It would be better to use a smaller resistor, more like 1/10 Ohm. Working at higher currents like you are, the errors introduced in the measurements will be smaller.
     
  15. Nov 18, 2011 #14
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    I never timed it I am just knew it was way way faster. This still does not answer my original question.
     
  16. Nov 18, 2011 #15

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Could you please re-state your question, now that you know V=I*R? Thanks.
     
  17. Nov 18, 2011 #16
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    I meant to say v/r=I still would equal 12 amps.
     
  18. Nov 18, 2011 #17

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    12V / 1 Ohm = 12A correct.

    12V / 2 Ohms = 6A correct.

    and so on... The higher the resistance put across a voltage source, the lower the current that flows through that resistance.
     
  19. Nov 18, 2011 #18
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Ok what if my battery was only producing 6A then I apply a 1ohm resistor to it will it increase my Amp to 12?
     
  20. Nov 18, 2011 #19

    berkeman

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    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    Please stop that. As mentioned already in the thread, a battery is a voltage source. It is not a current source. The fundamental thing produced by a battery is voltage. So when it us used in a circuit, the output voltage is the *cause*, and the current that flows as an *effect* is determined by the load resistance (or impedance) connected to the battery terminals.

    Please endeavor to understand the cause-effect relationship when dealing with voltage sources like batteries.
     
  21. Nov 18, 2011 #20
    Re: Please Help Me!!! THIS IS DRIVING ME MAD!!!

    This means if you draw 2.6A, the battery should last an hour.

    You are confuse between the battery storage capacity to the current. This amp hour give you the energy stored inside the battery. That has nothing to do with how many amp you can get out of it.

    For example a 12V carbon zinc battery, Lead Acid batteries of the SAME amp hour rating of 1 amp hour. If you put a resistor of 12 Ω across, you get 1A through on each and they all last one hour.

    But the two are very different. Lead acid has higher internal resistance and even if you short it, it can only give out ( just for example) 5A max. But the Lead acid is known to have much lower internal resistance, it can give out 200A. So if you put a 1Ω across them, the lead acid battery can give out 12A and it only last 1/12 hour. The carbon zinc battery can only give out 5A, so the voltage across the resistor will go down to 5 volts. This mean the output of the battery will drop down to 5V. The 7V that disappeared is dropped inside the battery's internal resistance. BUT it will last longer than the lead acid battery with the 1Ω.


    I hope this clarify for you. You are mixing up the energy storage capability of the battery with the peak output current.
     
    Last edited: Nov 18, 2011
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