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This is probably a really simple question but for some reason, I'm finding it difficult to grasp at an atomic level. My physics textbook has a statement that says:

"As the voltage across a component is increased, the current in the component also increases"

This is the basis for the Current-Voltage (IV) graph seen below:

enter image description here

However, I don't understand how we are supposed to increase the potential difference across just one component. Furthermore, I don't get why increasing voltage across the component should increase the flow of charge (current).

So my 2 questions are:

  1. How do we increase the voltage across just one component without increasing resistance?
  2. Why does increasing voltage mean there has to be an increase in the flow of charge (current)?
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  • $\begingroup$ en.wikipedia.org/wiki/Drude_model $\endgroup$
    – Jun Seo-He
    Jun 28, 2021 at 20:25
  • $\begingroup$ Do you know the definition of potential difference (voltage)? $\endgroup$
    – Bob D
    Jun 28, 2021 at 20:28
  • $\begingroup$ No. Not in much detail. Can you explain it to me? $\endgroup$
    – Anonymous
    Jun 28, 2021 at 20:30
  • $\begingroup$ As far as I'm aware, the voltage across a component is just the energy lost by every coulomb of charge. $\endgroup$
    – Anonymous
    Jun 28, 2021 at 20:32
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    $\begingroup$ I'm 15 so I haven't exactly jumped into that end of the pool yet. This is just something in my high school physics book that wasn't really explained. IK that voltage across a component is the voltage taken by the component per coloumb. But I don't see how increasing voltage, essentially increasing the electric p.d of charge increases the speed of the charge. $\endgroup$
    – Anonymous
    Jun 28, 2021 at 20:53

1 Answer 1

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  1. If you have a simple system with just one component, then increasing the voltage across the component is simply a matter of changing the voltage in your source. If your system is just a battery and a resistor, by replacing the battery for a stronger one, or two batteries in series you would increase the total voltage across the resistor. If you want more fine tuning of the voltage, there are power supplies in which you can regulate the output voltage.

  2. The voltage or more correctly, the voltage difference across a component is proportional to the energy a charge would gain or lose by traversing that component. There are many ways to explain why a voltage difference generates a current.

A simple way to understand this is that in physics, the systems always try to minimize the energy, which in this case means carrying the charges to a point of lower energy (across the component), generating a current.

A more detailed way would be to know that a potential difference causes an electric field to be present inside the component, and that electric fields generate a force on the charges inside the component, accelerating it.

One then could question where the kinetic energy gained by the charges go. The answer is that when the charges are traversing a resistor they "colide" with the atoms inside it, dissipating heat. That is why resistors heat up when you put a current through them. A mathematical way to describe this is through the Drude model, mentioned by @JunSeo-He.

More detailed models of this exist, but I hope these explanations will cover your basic doubts.

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    $\begingroup$ Thanks a lot. This explanation was very very helpful in my understanding of voltage. Does this electric field exist in all components? $\endgroup$
    – Anonymous
    Jun 28, 2021 at 21:16
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    $\begingroup$ Also, do you have any sources or books I can follow in order to obtain knowledge on the nature of current, voltage and resistance? $\endgroup$
    – Anonymous
    Jun 28, 2021 at 21:23
  • $\begingroup$ Any books, papers or websites will be very helpful $\endgroup$
    – Anonymous
    Jun 28, 2021 at 21:23
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    $\begingroup$ Thanks a lot for trying. I'll check their videos out. $\endgroup$
    – Anonymous
    Jun 28, 2021 at 21:56
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    $\begingroup$ I'm running out of ways to say thank you. This is extremely helpful. Wow. I appreciate it a lot $\endgroup$
    – Anonymous
    Jun 29, 2021 at 0:04

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