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More specifically, why is R=V/I. I understand what current is; the amount of charge that passes through a point in one second or Q/t. Voltage I'm not as sure, but from my knowledge its the amount of Electrical Potential Energy required to move a single unit of charge from one location to another. What I don't understand is how the resistance, a measurement of an opposing force to electrons, is equal to voltage/current. So in simple terms, why is this case?

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    $\begingroup$ Resistance is not a 'force' $\endgroup$
    – Aditya Prakash
    Commented Nov 28, 2019 at 17:49
  • $\begingroup$ Poor choice of word maybe. Or perhaps I'm just plain wrong. But at the end of the day, it resists flow, meaning that it has its own definition; other than being just defined by V/I. $\endgroup$
    – yolo
    Commented Nov 28, 2019 at 17:53
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    $\begingroup$ maybe this will helo ? hyperphysics.phy-astr.gsu.edu/hbase/electric/…. $\endgroup$
    – anna v
    Commented Nov 28, 2019 at 17:54

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Try to view the potential difference ($V$) as the work done by the external source and the current ($I$) as the speed with which you are driving or pushing the electrons. It is always easier for people to visualise stuff by using connections with mechanics. So if you want to push the electron with a certain speed then the work done by the external source increases if you increase the resistance (which is quite intuitive). And moreover the change is linear (which might not really be obvious at the first sight). Now try to use this analogy and see what happens if you vary the current while keeping the potential difference constant.

CAUTION :- This analogy works well for almost all the basic principles of electricity, but it would be better if it is not applied everywhere. This analogy is just a stepping stone to understand other advanced concepts. But in reality, this description is not accurate. This answer has been written only to provide an intuition about electric current. If you want a more rigorous treatment, then the link mentioned (by anna v) in the comments section of your question might be helpful.

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Think of this problem as a real life problem.

Suppose there is a job opening with a salary of one million dollars (The salary is analogue to the Voltage V ). Just like higher the voltage the higher the current we expect to flow, higher the salary higher the number of applicants for the job. If there are absolutely no qualifications required for the job, then every single person would apply for the job. Just like that if the wires do not have any resistance, we expect the current to be very large (i.e. infinite. okay may be not infinite but it will be as large as it can be). Whereas the specific qualifications required for the job will greatly decrease the number of applicants. In the same way the resistance in the connecting wires will hinder the electron flow and limit the current. So the relation between resistance(R) and current (I) is as:

                 I ~ 1/R

Also as we expect higher current with higher voltage

                 I ~ V

Combining the above two relations we get

                I ~ V/R

So

                R ~ V/I
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It might be easier to look at it as I = V/R.

Voltage is the amount of energy per unit charge. Current is the amount of charge that flows per second.

If you reduce resistance, the same amount of energy results in more charge flowing. In other words, lower resistance means electric charge is easier to move.

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    $\begingroup$ Doesn't this answer merely describe the effect of Ohm's law, rather than explaining the underlying logic that allows one to reach it? $\endgroup$
    – yolo
    Commented Nov 28, 2019 at 18:09
  • $\begingroup$ Pretty much. I thought that was what you wanted as you said you wanted a simple answer along the lines of what you said you understood about voltage and current. $\endgroup$
    – user68014
    Commented Nov 28, 2019 at 19:02
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    $\begingroup$ @yolo, there was no logic behind it. It's an observational law. Ohm measured the response of different conductors, and found experimentally that if he increased the applied voltage, the current increased proportionally. $\endgroup$
    – The Photon
    Commented Nov 28, 2019 at 19:19

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