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Homework Help: Find value of resistance to get max power in circuit

  1. Aug 2, 2006 #1
    Dear Sir,
    What is the method of finding the resistance value in a given circuit when one value of one of the resistance.Given E (EMF),one resistance in circuit, value of second resistance to be worked out to get maximum power.

    I tried to find as below
    min resistance will give max current.
    I^2R will be max
    as resistances are in series, for min equivalent resistance, external resistance has to be minimum.
  2. jcsd
  3. Aug 2, 2006 #2


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    According to the formula you gave, with zero resistance the power will be zero.
    Hint: express the current as a function of the source EMF, source output resistance and load resistance.
    Then express the power delivered to the load as a function of the current and the load resistance. Since you have already an expression for the current, you get an equation with only one unknown: the load resistance.
    The maximum power can be calculated by taking the derivative of the power relative to the load and making it equal zero.
  4. Aug 2, 2006 #3
    Maximum power is transferred when the impedance of the load equals the complex conjugate of the source. You may have to simplify the circuit somewhat to determine the source impedance.
  5. Aug 2, 2006 #4
    SGT Sir,
    Can you please calculate and show how to find out value of R by derivative method?
    Hammie Sir,
    This seems to be a different approach.What is impedence?Please explain your method to find value of "resistance.
  6. Aug 3, 2006 #5


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    Gold Member

    It depends what you mean. I think you want the maximum power in the unknown resistance, not in the known resistance R. If you wanted the maximum power in the known resistance R, you are right that the second resistance should be zero. But that's not the question.

    Let me call [itex] R_{unk} [/itex] the unknown resistance that you are looking for and simply "R" the known resistance. Then the current is
    [tex] I= {V \over R + R_{unk}} [/tex] where V is the emf of the battery.
    Then the power in the unknown resistance is
    [tex] P_{unk} = R_{unk} ( {V \over R +R_{unk} })^2 [/tex]
    You see that if [itex] R_{unk} [/itex] is zero, there is no power (obviously...the current is maximum but the resistance is zero so it can't dissipate any power). If [itex] R_{unk} [/itex] goes to infinity, the power again goes to zero (this time because the current goes to zero).

    So you need to find the value of [itex] R_{unk} [/itex] that will maximize the above expression. Just take the derivative with respect to [itex] R_{unk} [/itex] , set this to zero and solve for the value of [itex] R_{unk} [/itex].

  7. Aug 4, 2006 #6
    Thank you Sir.
    Actually the power is required from the whole circuit and not from the unknown resistsnce.
    Will you please help to get the answer for value of unknown resistance for getting max power?
  8. Aug 4, 2006 #7


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    Staff: Mentor

    The traditional way that this question is presented is to find the load resistance that maximizes the power transferred from a power source (with its output impedance defined) to the load. Are you sure that isn't what your question is asking? Can you post the exact wording of the question?

    Even if you are given a problem with one load resistor and you are supposed to calculate the value of a second load resistor in series with the first (to maximize the power transferred to the series combination of both load resistors), you will need to know the output impedance of the power source in order to make the calculation. The output impedance of the power source cannot be ignored.
  9. Aug 4, 2006 #8


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    I doubt that this is the real question. If it is, you are right in thinking that the maximum power is delivered to the source's output resistance when the load resistance is zero, but this makes no sense.
    nrqed has already done most of the work in providing you with the function relating power to the unknown resistance. All you have to do is taking the derivative and making it zero.
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