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Electric Potential Energy Conversion

  1. Mar 23, 2015 #1
    Hello everyone,

    Imagine two points that have a Electric Potential Difference of 10V and they are connected through a conductor wire. Let's call these points A and B:

    wire.png

    Now imagine we have a charge on A of 1C. To move the charge to B, we need to make a work of 10J, right?

    So, now, imagine the charge is on B. This means that the charge has 10J of Electric Potential Energy. Now, we let the charge goes through the wire, naturally. When the charge reaches point A, the potential energy will be reduced to 0, right?

    The electric potential energy was converted to what? Kinetic energy? heat energy?

    Now, what bothers me so much. Imagine now that we put a resistor between A and B:

    wire2.png

    Imagine the resistence of the resistor is 10ohm, for example. The wire is a good conductor, so the resistence is almost 0.

    Again, imagine 1C going from B to A.

    In this situation, the voltage drop of the resistor would be almost 10V, right? This is because the potential difference between B and A is also 10V and the resistence of the wire is almost 0ohm.

    In the picture A a charge that travels between B and A "used" 10J of energy to do it. Regardless of how the energy was converted, it was done. Going from B to A make the charge to lose 10J of potential energy.

    In the picture B, there is a resistor between A and B. The resistor consumes almost 10J of energy. This is ok, since the resistor is naturally an energy-absorber - it will transform the electric energy into heat energy or whatever. But what I can't understand is: How now when the charge travels through the wire it almost doesn't lose electric potential energy? The path between A and the negative terminal of the resistor have a voltage drop almost imperceptible, the same with the path between the positive terminal of the resistor and B.

    In the first scene, the charge lost 10J of potential energy just passing through the wire, without any resistor. In the second scene, only because a resistor was added to the circuit, all the voltage drop was directed to the resistor, and the wire became almost insignificant. Could someone explain me why this happens?

    If you guys could point all errors I made here and explain my question, I'll be so thankful. Thank you very much
     
  2. jcsd
  3. Mar 23, 2015 #2
    In the first case, in order to maintain a potential difference of 10 V you will need a huge current going already through the wire. This will dissipate a lot of heat in the wire.
    Any coulomb transported by this current will dissipate 10 J of energy by moving from one point to the other. The potential energy is converted to heat (which is the kinetic energy of the thermal motion of the lattice).
    In the second case you will have a smaller current through the resistor. But again each coulomb of charge transported will dissipate 10 J of energy. It s just that it will take longer to have 1C of charge moved from one point to the other. The electrical potential energy is again converted to heat.
     
  4. Mar 23, 2015 #3
    @nasu Thanks very much for the answer.

    Ok, this is a point that I'm not 100% sure. The voltage unit is Volt and a volt is 1J/1C. So, the voltage between two points tells us how many potential energy you will lose/gain by moving one coloumb between these two points. Since the voltage definition is with regards to only one coloumb, changing the voltage will not necessarily change the current, right? For example, if I two points have a voltage of 20V and I change it to 40V, I'm only changing the potential energy per coloumb (before 20J, now 40J). This will not make a bigger current flow between the points, it will just make the each coloumb flow with more energy.

    Thanks, this is what I expected

    Ok, I get it, this is also what I expected. But my question is a little deeper: Why the voltage drop is 100% concentrated on the resistor? For example: In picture B, why when the charge travels between B and the positive terminal of the resistor, it almost doesn't lose potential energy? All the voltage drop is done when the charge passes through the resistor. It's kind of weird. Picture A and B have the same wire. However, in Picture A, the charge loses 10J only by passing through the wire. In Picture B, the wire is almost insignificant, only because there is a resistor in the middle of the circuit. This is what is not clear to me: why this happens.
     
  5. Mar 23, 2015 #4
    You still don;t understand that in order to maintain that potential difference while a wire connects the points you need an almost infinite current through the wire.
    If you increase the voltage, the current will increase too. It seems that you mix two situations:
    1. Two points in space without an electrical connection, when indeed doubling the voltage you just change the potential energy. There is no other energy to speak of.
    2. Two points connected by a wire. If it's an ideal wire, the two points will have the same potential no matter what you do. If it's a real wire, with non-zero but very small resistance, you will have a current. This current will depend on the potential difference between the wires. The definition of the unit for potential is irrelevant. See Ohm's law.

    This is why the two cases are different. In the first one you have an almost "infinite current" through an almost zero resistor (wire). In the second you have a finite, low current through the same almost zero resistance (wire). They are far from being same thing.
     
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