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So I am having some issue understanding what exactly the voltage from an EMF in a circuit is exactly referring to. To start im guessing my confusion comes from the understanding of electric potential in the electrostatic sense. In electrostatics we basically say that electric potential energy of a charged particle in an electric field is converted to kinetic energy of a charged particle if let go from a particular point in that field. Since that charged particle gains kinetic energy and its mass stays the same it must gain a speed. So Electric potential energy in this sense gets converted to the speed of a charged particle. However in a circuit this doesn't necessarily work since the speed of particles /current is the same throughout the circuit. So what exactly is the electric potential energy getting converted to if its not kinetic energy? Is the potential energy in a circuit the potential for the kinetic energy in the particles of the current flow to convert to heat/light? And how is it possible that in a simple circuit with only a resistor that the voltage will drop to zero after the current passes through that resistor if the particles in that current flow still have kinetic energy?

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  • $\begingroup$ What does "the voltage from an EMF" mean? $\endgroup$ Mar 29, 2017 at 20:50
  • $\begingroup$ Electro Motive Force E.G. a battery $\endgroup$
    – XSoloDolo
    Mar 29, 2017 at 20:53
  • $\begingroup$ With regards to your last question, the kinetic energy doesn't have anything to do with the voltage applied to the charges. A voltage causes change in kinetic energy due to some electric force applied $\endgroup$
    – user86425
    Mar 29, 2017 at 21:07
  • $\begingroup$ albeit, there is no actual change because of the resistive forces in the wire $\endgroup$
    – user86425
    Mar 29, 2017 at 21:08

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The question is ambiguously worded. But I'll try to answer.

EMF is the total work done in taking a charge around the entire circuit, ie through the inner circuit in battery as well as through the external circuit.

Potential has nothing much to do with kinetic energy in the question.

The potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinite distance to that point.

The current drawn in the circuit is $I=\frac {V}{R} $ (Ohm's Law). There is a potential difference across the resistor. That is why the current passes through the resistor from higher potential to lower potential.

Informally, you can take it like this: There is a voltage drop across the resistor as it uses up current, so current floes from higher potential to lower potential.

Whatever I've written is as per current electricity. Electrostatics has no work in the question.

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  • $\begingroup$ I'm still a little confused then because isn't the current the same before and after it passes through the resistor? $\endgroup$
    – XSoloDolo
    Mar 29, 2017 at 21:27
  • $\begingroup$ You're forgetting the fact that voltage is a relative measurement. The "positive" terminal of a battery has a positive voltage relative to the other terminal. If measured the other way around the voltage is negative. After going through a resistor, the voltage drops from that of the first terminal to that of the second ("zero"). $\endgroup$
    – hdhondt
    Mar 29, 2017 at 22:32

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