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How does a battery send current through a wire?

  1. Sep 7, 2015 #1
    'The ends A and B of a potentiometer wire are connected a driving circuit consisting of a strong battery, a plug key and a rheostat. The driving circuit sends a constant current i through the wire AB.'
    Can anybody please explain me as clearly as possible what is meant by driving circuit and how does it send a constant current through the wire....... Please.......... I have read that battery creates a potential difference between the two ends and the free electrons in the wire more in a direction opposite to the field created by the p.d. Is it true ???
     
  2. jcsd
  3. Sep 7, 2015 #2
    Yes it is true. As for driving circuit, im just assuming here, but i suppose its referring to the fact, when you use a potentiometer, you have two sort of sub circuits( bad terminology). One consists of the a cell of known emf and , if you wish, some resistances etc. The other bit, is a cell, usually of unknown emf( maybe known, depends on what you're using the potentiometer for) with a galvanometer and a jockey. You move the jockey along the wire AB until the galvanometer gives a null defelction, the point along the wire where this happens is called the balance point. If you know the balance length( the distance from A to the balance point) you can calculate different things, which i wont go into here. The point is, the driving circuit refers, in this case, to the cell of known emf, the one without the galvanometer and jockey.
     
  4. Sep 7, 2015 #3
    Thanks a lot Ajay...... I have another doubt too.... 'The emf of a battery equals the potential difference between its terminals when the terminals are connected externally.' What will happen when something is connected externally ???
     
  5. Sep 7, 2015 #4

    sophiecentaur

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    The "NOT" is missing, there. The EMF will only be what's measured if there is no current taken from the terminals.
    When something is connected 'externally', current will flow. That current will be flowing through the internal structure of the battery, which is resistive. So there will be a voltage drop or voltage loss which is equal to Ir, where I is the current and r is this internal resistance. The more current you take (or the lower the external resistance`) the lower the voltage that you would measure across the terminals.
    btw, if you want to discuss a circuit, it is always better to actually produce a drawing -however simple - with labeled components, in the OP. Then we all sing from the same hymn sheet.
     
  6. Sep 7, 2015 #5

    ogg

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    If connected by a zero-resistance conductor, the electrons will flow and the battery will exhaust itself. (This ignores possible heating effects on the battery due to the short circuit condition). "Externally connected" just means "placed in a circuit"...and for almost all scenarioes a resistance will be present in the circuit. A battery is, on an abstract level, a system which has energy stored in it which when electrons are allowed to flow from the anode to the cathode (and the current flows from the cathode to the anode - blame Ben Franklin, he's the one (afaik) that made the electron negatively charged!) is able to "do work". Work is energy, of course. Potential really means the potential to do work. A voltage difference is a potential to do work. In order to measure voltage differences directly, this potential difference needs to do work on the potentiometer, which is why it must be connected to the circuit (measuring the voltage between those two points). Connect the potentiometer and the needle moves (or the LED display changes) which is work being done (or been done). The amount of work done better be very tiny compared to the energy available, or the measuring circuit becomes "part of" the driving circuit, and errors result.
     
  7. Sep 7, 2015 #6
    Thanks a lot sophiecentaur...
     
  8. Sep 7, 2015 #7
    Thanks a lot of ogg.....
     
  9. Sep 9, 2015 #8

    CWatters

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    Yes that's correct.

    If you connect an ideal battery to a resistor you have essentially built a circuit that tries to maintain a constant voltage across that resistor. Ohms law says the current I flowing in the resistor is given by

    I = V/R

    So if V and R are both constant then I is also constant.

    There are many ways to build a circuit that maintains a constant voltage across a resistor. You don't have to use a battery you can use a transistor or suitably configured amplifier. This is the way many constant current sources work. One such circuit is shown below..

    220px-Const_cur_src_112.svg.png

    This circuit maintains a constant voltage across R2. The way transistors work means the current flowing through the load is essentially the same as that flowing through R2. Perhaps revisit this when you have studied transistors.
     
  10. Sep 9, 2015 #9
    Thank you CWatters... :woot:
     
  11. Sep 9, 2015 #10
    Your question was "How does a battery send current through a wire?"

    If a wire is connected across a functioning battery the excess electrons on the negative terminal will be pushed by the charge on that terminal into the end of the wire connected to the negative terminal. Think of the analogy of a bunch of ping pong balls in a massively large diameter pipe. What happens is you shove a batch of balls (electrons) in one end and other electrons inside the wire are displaced down the length of the wire, finally pushing electrons out the end of the wire into the positive battery terminal. (actually all the way to the battery positive plate). The electrons that enter one end of the wire do not fly thru to the other end at the speed of light.

    As to potentiometers, there at least two types. One is a simple three wire device that allows you to pick off desired voltage by moving a slider down (around) a resistive divider. The other is a type of electrical test equipment (millivolt potentiometer) that allows you to measure rather low voltages without "loading" the source of the voltage and pulling the voltage down. It is pretty much obsolete now, having been replaced by digital millivoltmeters.

    Hope you are still around and that this is helpful.

    DC
     
  12. Sep 9, 2015 #11
    Thank you DC...:woot::smile:
     
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