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Electric Circuits

  1. Aug 14, 2009 #1
    I ran across this question on another site but I didn't understand the answer that was given.

    The question asked that if we have a circuit that connected the positive terminal of a 9V battery with the negative terminal of another 9V battery but not connecting the battery together, why would the circuit not work? Electrons doesn't go through the battery and the cations and anions that are needed to balance out the charges are present within each battery.

    Also, I'm wondering what negative voltage represetns in AC and DC circuits. For example, when I use my electric voltmeter and place the black line (normally placed the negative terminal) on the positive terminal of a battery and using the red line and placing it on the negative terminal, I get a negative voltage reading that is equal in magnitude to the voltage of the battery. I understand that in AC circuits that negative voltage means that the voltage is heading in opposite directions. But the thing is in a DC circuit a negative voltage means that the circuit won't go. Why is this? Are these two negative voltages taking about different things?

    I just graduated high school so I would appreciate it if you can answer my question in a simple fashion

    Thanks for any help that you can provide.
     
  2. jcsd
  3. Aug 14, 2009 #2

    cepheid

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    The battery voltage is quite literally the difference in electric potential between the two terminals of the battery. Let's say, for example, that we have a 9 V battery. What this means is that the electric potential of the + terminal is 9 V above the electric potential of the - terminal (when the battery is fully charged).

    Now, because of the way the voltmeter works, it always tells you the difference in potential between the red lead and the black lead. Therefore, if you place the red lead on the + terminal, and the black lead on the - terminal, then the red lead will be at a potential that is 9 V above the potential of the black lead. Hence, the difference (red minus black) will be + 9 V.

    Conversely, if you place the red lead on the - terminal and the black lead on the + terminal, then the red lead will now be at a potential of 9 V below the black lead. Hence, the measured difference (red minus black) will now be -9 V.
     
  4. Aug 14, 2009 #3

    cepheid

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    Not really. As I described, voltage is not a "thing" that "goes" anywhere (perhaps you were thinking of current?). Voltage is a difference in electric potential between two points in space. In an AC circuit, these potentials vary with time, in such a way that the voltage (potential difference) osciallates sinusoidally.

    No, it doesn't. Besides, what is positive relative to one reference point could be negative relative to another reference point, as illustrated in the example below.

    Let's say you first measure the battery voltage by placing the red lead on the + terminal and the black lead on the - terminal. Then you do it a second time, switching which lead is on which terminal. You get different answers because you changed what you were measuring:

    In the first instance, you were measuring the potential of the + terminal relative to the - terminal. This is positive.

    In the second instance, you were measuring the potential of the - terminal relative to the + terminal. This is negative.
     
  5. Aug 14, 2009 #4

    cepheid

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    The answer to your question is, WHAT circuit? You haven't made a complete circuit. There is no closed conducting pathway for charges to flow along. You have two batteries in series, that are ready to be connected to something. But so far, they haven't been.
     
  6. Aug 14, 2009 #5
    I dont think I quite understand the bolded...
    But, If Voltage exists, that means there is a potential difference between the terminals.
    If there is a potential difference, a circuit will have current "pushed"
    through it as long as the circuit is complete. It does not matter if you
    assign - or +. Current will flow. It might mean that the electric motor
    that you hooked up to the battery runs in reverse from what you expected.
    But if you switch the leads, the motor reverses. I used a motor so that you
    would realize that the current would switch directions if you reversed the leads.
    A light bulb would not lead to this realization unless it was an LED that allows
    current to flow in only one direction.

    I hope I understood this properly and did not confuse you.

    As an aside, if you could swith the leads on the battery 60 times every second
    you would have a type of AC current from a battery. Maybe this helps.

    oops I just read the above... I may have repeated some ideas.
     
    Last edited: Aug 14, 2009
  7. Aug 14, 2009 #6
    Thanks for the replies

    But in that case can you explain what negative voltage really is? I have a lot of misconceptions about negative voltage. Right now I think negative voltages is like those in electrolytic cells, which is voltage that must be added to make something work but I know this concept is wrong; however I've yet to find a good explanation on what it is. Does current still flow when we have negative voltages in AC and in DC and how does it differ from postiive voltages?

    Here is the circuit diagram that someone who also asked this question drew:

     
    Last edited: Aug 14, 2009
  8. Aug 14, 2009 #7
    If the voltmeter is hooked to a potential diff. (battery). If it is a needle voltmeter the needle will deflect in opposite directions if you change the leads to the battery. On a digital, it will read the same voltage, just + or -. The casing inside flashlights contain contacts that fit the post and the flat end of the battery. If you put the batteries in the wrong way you wont make contact with the metal contacts in the flashlight. So you have to put them in a certain way so you have a complete circuit (because you will make contact with the metal parts in the flashlight.Yes it would be more convienent. But my take is making the contacts so they fit certain terminals of the battery (post v. flat side) insures good contact. If both contacts in the battery were flat, and both sides of the battery were flat, you might not get certain contact. Thats why lots of battery operated devices have springs with contacts. The springs "push back" to ensure good contact.
    My take anyway.

    Oh btw, I find electrochemistry rather confusing. Voltaic stacks make sense to me, but the batteries today seem complex. So I just sit on the fact that batteries contain a potential diff. without delving into the chemistry as deeply as you are. Diff types of + and - ions in diff. batteries moving about might create confusion with the designated sign for voltage and terminals and such.
    I humbly await the day someone can come up with a battery that can store energy in a relatively small volume for a long time. Hydrocarbons do a wonderful job of storing energy, it would be great if a batteries could do the same. In Texas, they actually shut down wind turbines because they dont have enough transmission lines to get the electrical energy to houses, etc... You cant store that energy, you have to use it pronto.
     
  9. Aug 14, 2009 #8
    That's right. If a negative charge is put into a region of space where an electric field exists, if it's near the negative terminal then it has a more NEGATIVE POTENTIAL but a GREATER POTENTIAL ENERGY. The negative charge will then move spontaneously toward the HIGHER potential, and when it does it will lose POTENTIAL ENERGY (like a falling ball in a gravity problem).
     
  10. Aug 14, 2009 #9
    In each one of the batteries, the chemical reaction requires that electrons enter one side and leave the other side. Otherwise, there will be no net chemical reaction in that battery.

    Your diagram didn't get drawn because you put a backslash in your QUOTE tag.
     
  11. Aug 15, 2009 #10

    Dale

    Staff: Mentor

    You actually will get a very small amount of energy out of this arrangement. What will happen is that the half-cell reactions will occur at each electrode. As this happens each electrode will acquire a net charge. This net charge will then prevent the half-cell reaction from further continuing due to simple electrostatic repulsion. It does not take very much charge at all to do this. The only way for the electrode half-cell reactions to continue is if there is some mechanism to continuously "drain" the excess charge. This is done by a net transfer of some positive ions inside the battery in the same direction as the net transfer of electrons.

    In other words, in normal operation of a battery not only is there an electronic current through the wires of the load, but there is also an ionic current through the electrolyte of the battery. Both currents are essential for proper operation of the battery in order to prevent the electrodes from becoming charged (which would stop the reaction). Your arrangement blocks the electrolyte current.
     
  12. Aug 15, 2009 #11
    Thanks for the reply

    Can you explain how a net charge builds up in each battery (or the two half cells on separate batteries) and why a repulsion force would accumulate?

    Thanks
     
  13. Aug 15, 2009 #12

    Dale

    Staff: Mentor

    You have a current (briefly) going from one half-cell to the other, so charge accumulates by dQ/dt = I. And there is a repulsive force simply by Coulomb's law due to the accumulated charge.
     
  14. Aug 15, 2009 #13
    But how does both batteries get a net negative charge. If the batteries were initially neutral in the first place, then shouldn't one get negative and the other get positive? Thanks
     
  15. Aug 15, 2009 #14

    Dale

    Staff: Mentor

    Yes. Remember there are different reactions at the anode and the cathode.
     
  16. Aug 15, 2009 #15
    But then how can the electrostatic force be repulsive if the charges are unlike? Or do I have a misconception?
     
  17. Aug 15, 2009 #16

    Dale

    Staff: Mentor

    Yes, you have a misconception. Whether or not the charge on the anode is attracted to the charge on the cathode is irrelevant. What is important is if the charge on the anode repels the oxidation reactants away from the electrode and if the charge on the cathode repels the reduction reactants away from the electrode.
     
  18. Aug 15, 2009 #17
    So with a zinc- copper voltaic cell, in the cathode of one battery the Cu2+ is repelled and in the anode of the other battery the NO3- is repelled?
     
  19. Aug 15, 2009 #18

    Dale

    Staff: Mentor

    I don't remember the specifics of zinc copper cells, but that is the basic idea.
     
  20. Aug 16, 2009 #19
    Thanks for the reply

    My dad tried to explain this to me but I'm not sure if it's right since he hasn't touched this for 20+ years.

    He said that, in a copper - zinc cell with NO3- ion as the anion, the NO3- ion would repel the electrons at the cathode because the presence of electrons produces an excess charge (as the electrons don't originate in the same battery, they are extra). But my question is, if there is an excess negative charge then how come zinc ion isn't attracted from the anode to the cathode half cell to balance things out? Is it because the anode is balanced already and wouldn't move? Also how come the NO3- ion doesn't repel the electrons all the way back but just enough so that it stays on the electrode?

    Thanks
     
  21. Aug 16, 2009 #20

    Dale

    Staff: Mentor

    OK, I read up again to refresh my mind. Here is what happens:

    1) Some zinc metal leaves the electrode and dissolves into the solution, increasing the amount of zinc ion in solution and leaving electrons in the electrode.

    2) The electrons flow to the copper side.

    3) They attract copper ions in the solution which electroplate onto the electrode, reducing the amount of copper in solution and using up the electrons.

    4) Now the zinc solution has a net positive charge which pushes zinc away and thus keeps any more zinc from going into solution. Similarly the copper solution has a net negative charge and pulls the copper back and thus prevents any more copper from plating onto the electrode. The reaction stops.
     
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