Can anyone tell me the relationship between work done and potential difference??

mutineer123

CAN ANYONE TELL ME THE RELATIONSHIP BETWEEN WORK DONE AND POTENTIAL DIFFERENCE??
If more work is being done across a component(for ex due to increased resistance), does the pd across it increase or decrease? I think it should increase, because pd is the work done per unit charge, so if the work done increases, so would the pd. AM I right?

danmay

Yes, it would. There may be complications in terms of defining a potential, whether the work done is conservative, etc., but in general yes. Remember, however, that this increase comes from whatever "thing" (in your case, some kind of resistance) that increases the work required to be done.

Ken G

Well, one must be careful. Usually when one is talking about the work done across a potential difference, one is talking about the work done by the force responsible for that potential difference. The work then shows up as a change in the kinetic energy of the system crossing that potential difference, as per the "work-energy theorem." So if that is the context of the question, note that including resistance does not change the work done by the force associated with the potential difference (that can never be changed, it is given by the charge times the potential difference), but it means that there is also a resistive force doing additional work (which would actually be negative work and would subtract from the kinetic energy). So the answer to the question depends on whether one is talking about all the work being done by all the forces (in which case the work done could be less if there is resistance-- as it would likely be a positive work done by the potential and a negative work done by the resistive force), or if it is just the work done by the force associated with the potential difference (which is independent of any resistance that might be present).

tonyjk

We can say that for example a generator makes a positive work and all components that consume "electricity" makes negative work?

danmay

With respect to whatever you're trying to move, it would depend on whether its kinetic energy increases (positive) or decreases (negative) as a result of any work done.

Similarly for resistance, depending on whether it's against or along the original potential. For a resistance (say friction) that's symmetric with respect to the original potential, it could either increase or decrease the "total" potential depending on how you're trying to move the test object. In that case, the potential is probably ambiguous, if given only position information.

Ken G

It depends on the frame of reference, but in almost all situations you would normally see, friction does negative work-- regardless of whether the work done by the potential is positive or negative. Also, no "potential' can be associated with friction, so it's best to keep the "potential" concept and the "friction" concept quite separate in your head.

mutineer123

I am just an A level student, so have not come across work energy theorem, or friction as resistance...let me show you a question where a application of my question(in MY level) is shown.
http://www.xtremepapers.com/papers/C..._w11_qp_12.pdf

question 35.
If i was right(in this post), then my answer would have been C, but apparently the right answer is A, and I dont know why.

willem2

As the resistance goes up, the potential difference across it increases, but the current through it will decrease, because the total resistance increases, so in some cases
the work done can decrease, even if the potential difference increases.

total resistance: R+2 (series resistances)

current = 12 / (R+2) (ohms' law)

Potential difference across resistor = 12 R/(R+2) (ohms' law again)

work done = 144 R /(R+2)^2 (P = IV)

This function has a maximum.

Ken G

That's correct, and if you want a more physically intuitive version, the point is that when the load resistance is much smaller than the internal resistance, the current is not affected much by the load resistance, so the power dissipated in the load resistor will be proportional to the load resistance R. But as the load resistance exceeds the internal resistance, now the internal resistance is what is negligible, and you just have a fixed voltage across the load resistance R, which by P=V^2/R means the power will drop like 1/R.

danmay

Negative work doesn't necessarily mean a more negative potential. It depends on the nature of resistance/friction. It also helps to know why friction and potential are usually kept separate (look up Wikipedia or other sources for the complete explanation); but yes, they are usually not associated with one another.



willem2 (#8) and Ken G (#9) have got it.