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An annoying voltage question

  1. Feb 16, 2009 #1
    You have the following:
    1.Two batteries one of emf 9V the other of emf 12V.The batteries are ideal(zero internal resistance)
    2.Connecting wires of zero resistance
    2.An ideal voltmeter(infinite resistance)

    The batteries and voltmeter are all connected in parallel.What voltage will be measured when the batteries are connected positive terminal to positive terminal and negative terminal to negative terminal?What voltage will now be measured if one of the batteries is turned around to face the other way?

    (Dont get too annoyed if you are unable to come up with an answer)
  2. jcsd
  3. Feb 16, 2009 #2
    i'm going for the obvious and probably wrong answer:
    batteries ++ -- : (12v+9v)/2 therefore: 10.5v
    batteries +- +- : 12v+9v = 21v
  4. Feb 16, 2009 #3
    0V and 0V
    They have exploded.
  5. Feb 16, 2009 #4
    Well done Kittel Knight.
  6. Feb 21, 2009 #5
    It's not only the batteries that have exploded. The zero internal resistance of the wire has dissipated a power of V^2/R = (12 V - 9 V)^2 / 0 ohms = infinity watts. That's enough heat to vaporize the universe.

    You can get the same effect if you take an ideal current source, don't connect to anything, and turn it on. Pushing any required current through an infinite resistance, the power dissipated becomes infinity.
  7. Feb 21, 2009 #6


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    That's hardly dramatic. Most of the universe is vapor anyway!
  8. Nov 13, 2010 #7
    Hi Dadface - I respectively disagree with all of the answers. I believe that there is no way to really tell what the voltmeter will read given all of your conditions....because...let me give an example: You have a very large capacity battery and a very small battery, lets say a regular 12 volt automobile battery and a some little small 9 volt hearing aid battery. What is going to happen to that little battery when they are connected together, either way, postive to positive or negative to positive-the little battery will not be able to maintain it's internal resistance of zero and just open up, maybe with a bang. But, that large battery will be able to hang in there with just a little less voltage than it started with because it took some energy to blow open that little guy. So in this case, the answer would be, a litttle less than 12 volts. And of course if the 9 volt battery was large enough, and the 12 volt battery was some little thing and was opened, then the voltmeter would read something less than 9 volts. Being one to try not to say 'always' or 'never', because if I did I would probably be wrong, that is just the way I see it. I think that this is an example of the old adage that things don't defy physics, it is just a case of not considering all of the facts. If some disagree with this, that is okay, I have been wrong before. :)
  9. Nov 13, 2010 #8
    I too disagree with all the answers. These are ideal batteries and wires. They're allowed to conduct infinite current through zero ohms. The two ideal batteries connected ++ -- will always have the voltage of the lower voltage battery. Infinite current will flow from the higher voltage battery backwards through the lower voltage one.
  10. Nov 14, 2010 #9
    Hello YesIam and hello skeptic2.You are both right to disagree in that the circuit as described is not real.If R is truly zero then the current as calculated could be described as infinite but in reality that is meaningless and cannot happen.In short it was a thought experiment.
    I think the question is useful in that it is a good illustration to show that care should be taken when using simplifying assumptions.We might often ignore internal and other resistances to get answers that are approximately correct and good enough for the task in hand,but in this case we get no meaningful answer.
    Let's make the question a bit more real by giving the batteries internal resistances and to make the numbers simpler let the resistance of the 9V battery equal 1 ohm and the resistance of the 12V battery equal 2 ohms.With the first way of connecting the batteries the current will be 1A and the recorded voltage will be 10V(for the 12V battery 12V minus 2v dropped across its internal resistance and for the 9V battery,where the current is driven in the opposite direction to what it normally delivers,9V plus 1V dropped across its internal resistance).With the second way of connecting the batteries the current will be 7A and the recorded voltage 2V.
  11. Nov 14, 2010 #10
  12. Nov 14, 2010 #11
    Hi Dadface - First of all I believe that theoretical questions can have real useful answers - it is done all of the time. Secondly, I try to have an open mind when it comes to something new. Thirdly, I believe as I said, that when one uses, "always" or "never" in a serious sentence or 'implies same', is usually wrong. When a good instructor starts his class he will start with somthing like this, "We will learn a lot of facts in this class but we believe that half may be correct and half may be wrong...but...we don't know which half is which.
    Most of us in this area are between some kind of basic electronics school and/or has been a professor emeritus with 50 years of teaching experience in the hard sciences. And I believe the more you learn, the less you know and...no one knows but an almost infitesimal amount about anything. That last part is my quote. The 'stuff' you learn in school are just some facts to help you really learn after school. And the facts 'do not' have to be correct to help you in this endeaver. A good example is that 1 amp of current through 1 ohm of resitance does not drop 1 volt. Ohm's law is just a close approximation to help with the day-to-day needs. The real result changes from time to time-the last time I remember the error was in the third decimal place. But life goes on and this is no concern to 'most of us'. Fact is I have a good feeling that all formulas are only approximations-it is just the way formulas are derived. I would sure like to hear of a formula that claims to be exact.
    Back to the batteries...boy, I can't for the life of me get anywhere close to your explanation, I am very sorry...but lets try to hang in there.
  13. Nov 14, 2010 #12


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    Seconded. :smile:
  14. Nov 16, 2010 #13
    Power is not equal to heat. Heat potential of the battery is obviously limited to the amount of charge and the voltage on the charge.
  15. Nov 16, 2010 #14
    Regor60 - There is so much bogus 'stuff' on this forum you wouldn't believe. First it is unmoderated...this is okay, in a way. But even if it was moderated it wouldn't keep all of this bum dope from showing up. Someone can think, "Oh, I think it works this way, therefore it does-I am going to tell everyone." I just chanced upon this forum a few days ago and this is just my opinion now...you know everyone has one but I haven't seen anything that I think has much merit. Some of the things 'may be kinda straight' but usually it is more like, "This may be close to correct...sometimes." Unfortunately things will not change here because it is usually more like, "I have read it is like this or I think it is like this and I have a good mind...and it is made up.
    For example, now mind you I may be wrong and looking at it the wrong way, but lets look at the last thing you said in the above. "Heat potential...voltage..." Now let me give an example with a normal 12 volt automobile battery and give it's output capacity in coulombs. Do agree that this can be converted to 'power' or 'heat' in any of a variety of units-a given battery with so much potential output. Now take this same battery and take out the cell separation walls, 'trying to' keep everything the same...let's forget about the space that was used for the cell walls. What do we have now-a 2 volt single cell, or a little more. But how about the potential power or the potential heat output of this new 2 volt single cell. In short, I will maintain that they will 'roughly' be the same for both batteries. So...I maitain that you can leave the voltage out the equation and just put a period after, "...amount of charge." As for your statement of, "Power is not equal to heat." - I have no idea what-so-ever where you are coming from on this. Years ago I used an hp Calorimeter many times to measure power out of various devices and this is exactly how this Calorimeter worked. Power was supplied to an insulated container of oil for a given amount of time and then the temperature of the oil was measured. Of course the temperature rise was a direct indication of the amount of heat that was transferred to the container of oil. Anywho...maybe you don't see it this way but life will go on-even with all of this bogus dope on this forum. Now I repeat, this is just the way I see it and every day something new comes along and disproves yesterday's ideas.
  16. Nov 16, 2010 #15
    Regor60 and others - Let me explain my above statements...quickly...quickly. As I had said I had just come across this forum and...had read only two or three threads. My only defense is, "...this is just my opinion now..." But I just had a chance to look at some of the other threads in other areas and WOW, I saw stuff that I could never dream of understanding and of course way, way above me. As I said, "No one really knows all that much about anything." And of course I sure humbly fall into that, "No one..." category Sorry for any confusion on this.
  17. Nov 16, 2010 #16
    I forgot one little thing about the battery - the internal connections have to be changed to put the six cells in parallel. Sorry...
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