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Voltage drop in a circuit

  1. Feb 17, 2015 #1
    So i'm having trouble visualizing how voltage drops more within different parts of a circuit, such as a resistor vs. wire. I know all the general equations but the concept is hard for me to comprehend and I am stuck on one notion. So, say you have a simple DC circuit with just a resistor and wire connecting the two ends. Since the resistor has much more resistance than the wire, the drop in potential will be much greater across the resistor than the wire. Consequently, since the drop in potential per unit length is higher for the resistor wouldn't that mean the electric field is greater in the resistor? but how could the electric field be greater and what causes the change in E field?
    Last edited by a moderator: Feb 18, 2015
  2. jcsd
  3. Feb 18, 2015 #2


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    Staff Emeritus
    Science Advisor

    If I understand it correctly, voltage drop is not a change in the electric field, it is a loss of potential energy.

    From wiki (underlining mine):
    Voltage drop describes how the supplied energy of a voltage source is reduced as electric current moves through the passive elements (elements that do not supply voltage) of an electric circuit.
  4. Feb 18, 2015 #3


    Staff: Mentor

    Yes, assuming the wire and resistor were equal in length.

    Assume the contrary. What would happen if the E field were uniform? Think about Ohms law, currents, and conservation of charge.
  5. Feb 18, 2015 #4
    Okay I think i'm starting to understand. so if the electric field were uniform then the current going the wire would not be uniform because it takes more force to move current through a resistor than the wires? and if the electric field were uniform then the change in potential energy would also be uniform. Since it takes more force, and therefore a greater E field, to move through a resistor then it should cause more change in potential energy. However instead of converting that electric potential energy into kinetic in the form of a faster current, it gets converted to thermal energy? Am I on the right track?
  6. Feb 18, 2015 #5


    Staff: Mentor

    You are getting closer, but please forget about electrons and kinetic energy when thinking about electric current. The flow of energy (I.e. Power) in an electric circuit is proportional to voltage drop times current. V*I. That is what you should remember.

    When the same current flows through the wire and resistor, the voltage drop across the ends of the wire is nearly zero, but the voltage drop across the resistor is not zero. Therefore, the power dissipated in the wire is zero*current and the power dissipated in the resistor is V*current.
  7. Feb 18, 2015 #6


    Staff: Mentor

    Yes, this is the key point. Think about what would happen if there were a non uniform current.

    If more current is leaving an area than is entering then that region would become progressively more negatively charged. That would produce an E field which would invalidate the original assumption of uniform E field. In fact, this charge accumulation would continue until the current became uniform and the E field non-uniform in exactly the fashion that you first noticed.
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