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Batteries(and v-sources in general) in parellel

  1. Feb 1, 2012 #1
    I've read that lets say 2, 9v batteries, in parallel will give you more power whilst keeping 9V.


    Isn't this an engineering "no-no"?

    Voltage sources have near 0 internal resistance, why would battery send current to a load, when it can nicely go through the other battery?

    Or will the fact that they have both + at the same side "repel" any current that tries to enter the other battery?
  2. jcsd
  3. Feb 1, 2012 #2
    Note the caveat in Fig 4 about parallel combination.

    Basically you are OK so long as the cell fails high resistance. This is the normal mode of ageing in a battery - its internal series resistance gradually creeps up over time.
  4. Feb 1, 2012 #3
    Wait so batteries, normal ones, don't have very low resistance?
  5. Feb 1, 2012 #4
    (steady state) Current only flows between points of different potential, so if the two batteries have exactly the same voltage, no current will flow from one to the other. Since in real life two batteries attached in parallel will not have exactly the same voltage, a current will flow to equalize the two, which is a function of the internal resistance and the small difference in voltage until they are exactly equal. This however represents one battery charging the other, which is not something you are supposed to do with normal non-rechargeable batteries. For rechargeable batteries I don't see a problem in doing this. (think of jump-starting your car)

    I think batteries in parallel is not a very good idea in general, but in specific cases it could be fine. Combining brand new batteries of the same type and brand should be ok, but I still wouldn't do it without some additional components to ensure that current is drawn from each battery equally.
  6. Feb 1, 2012 #5
    Take a D cell, a C cell, an AA cell and an AAA cell.

    They all have the same terminal voltage, but you could never pull an amp out of the AAA cell, the internal resistance is just too large. Basically the difference in capacity can be stated in terms of internal resistance.
  7. Feb 1, 2012 #6
    Interesting. I was actually asking, because I will need power for my RC chopper, and I was looking to keep the voltage but increase the power.

    Makes sense. Thank you.
  8. Feb 1, 2012 #7
    I think I understand. When I was little, I saw those big cells but they gave only 1.5 V or so, I was always saying: pff thats one bad battery, that BIG and so little volts :D
  9. Feb 1, 2012 #8
  10. Feb 1, 2012 #9
  11. Feb 1, 2012 #10
    The model of a battery is an ideal voltage source with a series resistance (an ideal voltage source is a hypothetical source capable of providing infinite current through zero resistance). As the energy gets used up, you can think of it as the series resistance getting bigger, leading to an observed voltage drop due to more and more voltage falling off across the increasing resistance.

    When you measure the voltage your are creating a circuit with a voltage divider made by the battery's resistance and the resistance of your volt meter. The voltage measured will indicate the battery's resistance as compared to your meter. As the battery gets used up, the voltage drops, not because the potential difference between the positive and negative terminal is changing, but because the current-providing capability of the battery is dropping, equivalent to that resistance increasing. Consider a nearly dead 9v battery, modeled as an ideal voltage source with a resistance comparable to your voltmeter, say around 10 megaohms, your attach your multimeter with an input resistance of 10 megaohms and you read 4.5 volts and you conclude that your battery requires replacement.
  12. Feb 1, 2012 #11
    I understand, I will try that. Thank you kind sir.
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