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I Decreasing resistance proof

  1. Apr 4, 2016 #1
    Is it always true that if I have any system of resistors and I calculate the resistance between two points, when I decrease the resistance of one resistor, then the resistance measured between the same two points as previously will not increase?

    I.E if i have lots of resistors between two points and I decrease the resistance of one of them, will the total resistance always not increase? If so, is there a proof?
  2. jcsd
  3. Apr 4, 2016 #2
    It depends on your 'system'.
    In the simplest case of two resistors in parallel, if you reduce the resistance value of one resistor, then the resistance of the combination also reduces.
  4. Apr 4, 2016 #3
    I'm wondering if it is true in general, no matter what the system is like, as long as it only has resistors? Perhaps someone has a counterexample or knows if this is true.
  5. Apr 4, 2016 #4
    Hmm, actually what I posted above is also true for two resistors in series, but the net effect is different.
    In general if you have a circuit and reduce the resistance in some part of it, more current will end up flowing through the circuit.
    Which is the same thing as saying that the resistance of the circuit as a whole is reduced.
  6. Apr 4, 2016 #5
    It does intuitively seem like that, but is there a proof?

    I could have a very long and complex system, and maybe if i decrease a particular resistance by a specific amount the currents will rearrange in such a way that the total resistance will increase. Is there a way to show that it cannot happen?
  7. Apr 4, 2016 #6
    You might be able to get the sort of result you're talking about if the circuit also included one or more transistors.
  8. Apr 4, 2016 #7


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    It is always true. Equivalent resistance will go down if and only if there is current flow through the resistor that reduces its value, and it will never increase.

    You can probably show that via star-mesh transforms using some derivatives with respect to the changing resistance.
    The special case of no current is trivial.
  9. Apr 5, 2016 #8
    That is a good idea, using the star-mesh transform we can always transform it to just four nodes and then by letting the computer do the algebra, we can solve for the total resistance of the general case with four nodes. Then I let the computer calculate the derivative (it was some horrible long expression) of the total resistance with respect to each resistance and it came out to be the square of something, which is never negative and therefore proves the statement.
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