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My question is to verify if my thought process below is correct.

So in a circuit the charge will flow (the current). When the charge flows there is resistance which is the collisions of the charge with the positive ions, this causes the charge to have to work(energy is transferred to component)to get through components. (hence potential difference is the work done per unit charge).

I think my thought process is incorrect because if I follow my thought process it does not match up with V=IR.

For example, when the resistance is increased the amount of current able to pass through components will be less (as the charge will take longer to pass through the component). And then the voltage will be higher because the charge has to do more work to pass through the component because there is more resistance.

So maybe my thought process is correct as the voltage increases with the resistance, but it doesn't work with my idea of current.

Please can you help clarify my understanding

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2 Answers 2

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For example, when the resistance is increased the amount of current able to pass through components will be less (as the charge will take longer to pass through the component).

The current will be less only if the voltage across the resistor is fixed.

And then the voltage will be higher because the charge has to do more work to pass through the component because there is more resistance.

No. The current is less for a greater resistance if the voltage is fixed.

You can think of the voltage as the amount of work per unit charge that the source of the voltage is capable of doing to move the charge between two points. The greater the amount of work that is needed to move the charge between the two points (the greater the resistance) the less charge per unit time (current) a fixed voltage source will be able to move between the points.

Hope this helps.

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  • $\begingroup$ Thank you for your answer. It has helped. I do have a follow up question. So are you saying the voltage across a component is always the same and that it does not change depending on changes to like resistance because we see voltage as the work per unit charge CAPABLE to move charge across the component. And I am guessing that this is where we get V = E ( electric potential energy in joules)/ Q (charge in coulombs). $\endgroup$
    – Muffin
    Commented Apr 19, 2023 at 21:40
  • $\begingroup$ Also is this way of thinking (seeing voltage as the CAPABLE work per unit charge) what happens in reality or this is a sort of compromise so that the explanation works. Because in my mind I think that when there is an affect (change in resistance) there will be an effect (change in voltage). Thank you for taking time out of your day. (I do feel a bit guilty that my questions are becoming annoying and repetitive) $\endgroup$
    – Muffin
    Commented Apr 19, 2023 at 21:45
  • $\begingroup$ I think we need to come to some common understandings. First of all, do you agree that if the resistance is doubled and that results in the current being halved, that the voltage across the resistor has to be the same? Yes or no answer, please. $\endgroup$
    – Bob D
    Commented Apr 19, 2023 at 23:59
  • $\begingroup$ Yes I do agree. $\endgroup$
    – Muffin
    Commented Apr 20, 2023 at 11:59
  • $\begingroup$ @Muffin “Because in my mind I think that when there is an affect (change in resistance) there will be an effect (change in voltage)”. There will be an effect if the voltage across the specific resistor is not fixed. For example, consider two equal resistors of value $R$ in series with a fixed voltage source of value $V$. Although the voltage is fixed across the series combination, the voltage on each individual resistor depends on the value of the other resistor. In this case, the voltage across each resistor is $V/2$ and the current in both will be $V/2R$. $\endgroup$
    – Bob D
    Commented Apr 20, 2023 at 13:08
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Your thought process is basically right except for where you say the voltage will be higher if resistance increases. That would only be the case if you somehow maintain the same current.

The charge carriers (typically electrons) in a conductor often bump into the atoms. Each of these interactions causes an atom to vibrate, raising the temperature of the conductor. In this way the kinetic energy of the moving charge carriers is converted to thermal energy of the material. The charge carriers' kinetic energy is continually being replenished by the electric field that drives the current. So the electric field does work on the charge carriers, but that work is dissipated as thermal energy.

If you consider constant voltage (as in a DC circuit) $V$ cannot change. So increasing $R$ reduces $I$. The voltage would need to increase if you wanted to keep the same current, but if the voltage stays constant the current will drop.

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  • $\begingroup$ Thank you for your answer and your explanation of heating in electric circuits. When you say the kinetic energy is transferred to the atoms causing vibrations and then heating; is this only kinetic energy or some electric potential energy can be transferred to make this effect. I am thinking it is the kinetic energy because it makes sense as it is constantly being replaced meaning it can continue to heat and heat but if it was electric potential energy it would run out of energy and have to be given some more at the source. $\endgroup$
    – Muffin
    Commented Apr 19, 2023 at 21:56
  • $\begingroup$ Also is that explanation to do with the electric field a sort of extension of the generator effect where we can have an induced current and induced potential difference without having a electric potential energy source but having an electric field and charge carriers $\endgroup$
    – Muffin
    Commented Apr 20, 2023 at 11:56
  • $\begingroup$ Energy: the distinction between kinetic energy and thermal energy is fuzzy at small scale. Thermal energy in a material is really just oscillations of a huge number of atoms. But all oscillations contain a mix of kinetic and potential energy. Picture an atom electron struck by a conduction electron. It goes far away from its nucleus, so that it gains PE and loses KE. When it falls back it loses PE and gains KE. When we consider one particle we talk about KE and PE, but when we consider 10^23 particles we call it thermal energy. It's the same concept, but we account for it differently. $\endgroup$
    – Rich006
    Commented Apr 21, 2023 at 11:32
  • $\begingroup$ Electric field: Any time there's a potential difference (voltage) there's an electric field directed from higher to lower potential. Potential is more of a conceptual convenience (a mathematical construct) than a real thing. It's the electric field that actually exerts force on the charge carriers (conduction electrons) to make them move in a preferred direction through the wire. This electric field can arise in 2 ways: (1) separated charge, as provided by a battery or capacitor; (2) a changing magnetic field as provided by a generator. $\endgroup$
    – Rich006
    Commented Apr 21, 2023 at 11:38
  • $\begingroup$ ok thanks for the insight $\endgroup$
    – Muffin
    Commented Apr 22, 2023 at 18:45

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