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Why does a battery die when connected on prallel?

  1. Feb 26, 2012 #1
    During one of my labs we were using circuits. I connected a 9V battery in parallel with a 2.kΏ resistor and a multimeter. I meaured the current across the circuit and noticed it went really high at the beginning (about 30mA) and then it slowly went down and the battery became really hot and unusable. Why did that happened? I've been trying to find why the battery would get hot and stop working. Any help?
     
  2. jcsd
  3. Feb 26, 2012 #2
    Because you created a short across the battery terminals.

    It would be like connecting a light bulb to the battery. Depending on the wattage of teh resistor, it may have gotten hot as well.
     
  4. Feb 26, 2012 #3

    DaveC426913

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    The battery is obviously shorted.

    What two devices were in parallel exactly? The resistor and the multimeter (as an ammeter)? What setting was the multimeter on?
     
  5. Feb 26, 2012 #4

    OmCheeto

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    It sounds like you had a short circuit.
    9 volts / 2000 ohms = 4.5 ma
    This would occur if you connected your multimeter in parallel with your resistor while in the amps mode. It should be placed in series with the resistor when measuring current.
     
  6. Feb 26, 2012 #5

    DaveC426913

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    Yeah. That's where I was going too.
     
  7. Feb 26, 2012 #6

    jim hardy

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    he did say ""I meaured the current across the circuit .."" and got 30 ma, presumably not through 2K..

    Polaris, go to "Electrical Engineering and read this thread :

    "" why must ammeter be connected in series?""
     
    Last edited: Feb 26, 2012
  8. Feb 27, 2012 #7

    NascentOxygen

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    The current was much higher than 30mA. You were not reading the right scale. I surmise that you were not using a digital meter? 300mA possibly.
     
  9. Feb 27, 2012 #8

    davenn

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    no, as Jim hinted at, I bet he was trying to read the current in parallel across the cct instead of in series and the poor meter was shortcircuiting the battery

    Dave
     
  10. Feb 27, 2012 #9
    I think the current would have to be much higher than 30 ma with a short circuit. That would mean the internal resistance is 330 ohm, and only 0.27 Watt of power would be produced in the battery.
     
  11. Feb 27, 2012 #10

    NascentOxygen

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    yes

    That premise was fundamental to my reply. That the meter was shorting the meter is not in question, a 9v battery is not going to get hot powering either a 330Ω load, or delivering 30mA.

    Not only was the meter incorrectly connected to the resistor, but OP has not interpreted the meter reading correctly, regardless of whether it was digital or analog. That 30mA was not 30mA, maybe it was 300mA.

    I suggested analog, simply because I find it difficult to see how a digital reading could be misread too low by a factor of 10. Maybe there was a blob of solder on the readout and this was mistaken for a decimal point?
     
    Last edited: Feb 27, 2012
  12. Feb 27, 2012 #11

    jim hardy

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    here's short circuit tests on the common 9 volt 1604 "transistor radio battery" . Looks like they are capable of ~ 4 amps, briefly.

    friedrichengineering.com/web_documents/9volt%20Battery.pdf
     
  13. Feb 28, 2012 #12
    Quite a few (cheap) digital multimeters have a separate socket for an unfused 10A range. You have to plug and switch range, so a misreading of a factor 10 (or even 100) is definitely possible.
     
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