Ohm's law's gives us two relations: $V\propto R$ and $V\propto I$. But why is that so?
Potential difference($V$) between two points depends on the concentration of charges between them. Current flows from a region of high concentration of electrons to low concentration of electrons(low to high potential). But I can't seem to figure out how $V$ is dependent on:
- Current($I$): According to my book-
Potential difference between two points is equal to the work done per unit charge in moving a positive test charge from one point to another.
So why does current, which is number of charges flowing per unit time a factor in the work done on unit charge? If the work done on a unit charge to bring it from one point to another is $x$ then the work done to move $5$ charges between the same points is $5x$, but the potential between the two points, as per the definition, remains $x$.
- Resistance($R$)-
Suppose there is unit positive charge Q which is being taken to a greater positively charged body.
Suppose external work done is $x$ and then some resistance is provided in the path of the charge.
We know that $$\Delta W_{external}=\Delta PE + \Delta KE + W_{other}$$ where $W_{other}$ comprises other forms in which energy is being lost, e.g., heat, against friction, etc. So isn't work done against resistance coming under $W_{other}$? Just like in gravitational potential energy, where if we lift a body up a certain height through air, then stop the ball and keept it at rest at that certain height, then $W_{external}=\Delta PE+ \Delta KE + W_{againstresistance}$. $\Delta KE$ is $0$, and the total external work done is $\Delta PE + W_{againstresistance}$, but the change in potential always remains $mg \Delta y$!
Then why is the potential difference not only dependent on the shortest distance between and the charge concentration of the two points, but instead on $I$ and $R$?
Best Answer
For many physics problems the way to solve them is to get the right level of abstraction. You have drifted into the wrong level.
You need to be careful about comparing currents in wires to motion of particles affected by potential differences on static charges. Your diagram shows a test charge being moved from infinity towards a fixed charge. Electric circuits are not like that. Ohms law is about circuits.
As a snap-you-out-of-your-rut consideration, consider an electron going around a circuit several times. It moves "up" the resistance each time. Does it therefore wind up at a higher potential each time? Or does something happen so that it gets back down the hill each time?
The potential difference between the leads of a resistor determine the energy a single electron will dissipate in the resistor as it goes through. It does not gain that much energy. On the other side of the resistor it will have pretty much the same energy it had in the wire before the resistor. It will be stumping along at the same speed it had before.
The electron is getting pushed through the resistor. The restance is using up the energy of that push, mostly as heat. Tiny bit of electromagetic radiation.
When you increase the potential, increasing V, more current flows. And each electron loses more energy passing through the resistor. Double V you also double I, and you multiply the power into the resistor by 4. They do go faster, but the kinetic energy in an electron is quite small compared to the energy the current is carrying.
So your diagram of bringing a test charge from infinity is misplaced. You want to think of the electrons on a string getting dragged across rough surfaces. They don't wind up with more energy even if you pull harder. You just get more per second going by any given spot. And by pulling harder each one gets more violently smacked around by the rough surface.