Find the resistance of the resistor

Discussion in 'Homework Help' started by Agonche, Aug 26, 2011.

  1. Agonche

    Thread Starter Member

    Aug 26, 2011
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    On a real voltage source of 10V with internal resistance of 100\Omega, a resistor is connected. The power on the resistor is 200mW. Find the resistance.

    [​IMG]

    I know Ohm's law, Kirchhoff's laws, Power formulas
    P=I^{2}R
    P=\frac{E^{2}}{R}
    P=EI

    I tried finding R or I from R=\frac{P}{I^{2}}


    This may be an easy example but I don't know I can't find R.
    I'm solving harder problems, but can't find this one.
    Thanks in advance.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The source terminal voltage

    V=10-100*I

    The load power

    P=0.2=V*I

    V=0.2/I=10-100*I

    and so on to solve for I from a quadratic equation & finally R=P/I^2=0.2/I^2

    There is an unexpected outcome by the way......
     
    Last edited: Aug 26, 2011
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  3. Agonche

    Thread Starter Member

    Aug 26, 2011
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    yeah, I tried this
    but from the quadratic equation there are two outcomes
    I1 and I2,
    0.028 and 0.072
    Im really confused.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply you have two proper solution to your problem.
     
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  5. Agonche

    Thread Starter Member

    Aug 26, 2011
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    @Jony130
    wow how can that be true.
    never thought there can be two solutions...
    I solved problems like this one before, but from the quadratic equation there was only one outcome.
    for example I remember this one...

    6I^{2}-24I-24=0
    this equation has only one solution, which is 2.

    anyway thanks Jony and t_n_k
     
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    It is not difficult to see that there must be two solutions to this problem. The maximum power will be transferred into the load when it is made equal to the generator resistance, in this case 100Ω. This results from the Maximum Power Transfer Theorem:

    http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

    The load voltage at maximum power will therefore be half the open-circuit value, i.e. 5V, so that the current will be 5V/100Ω = 50mA, and the output power will be 5V*50mA = 250mW.

    The load power will also reduce if its resistance is varied either up or down from 100Ω, so we can expect to obtain 200mW for two current levels, one somewhat greater than 50mA and another somewhat less.
     
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  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Plotting V-I characteristic against the P=0.2W curve shows there are two valid conditions for 0.2W load power.
     
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  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Just as an aside, this equation has two solutions, neither of which is 2.
     
  9. Agonche

    Thread Starter Member

    Aug 26, 2011
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    yep, you're right.
    +24 instead of -24, the "c" element.

    6I^{2}-24I+24=0
     
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