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Maximum power on a resistor

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the maximal power on resistor R.

    2. Relevant equations

    3. The attempt at a solution
    I used Thevenin theorem. I found that Z_t=5+j10 , and U_t= 10-j10. The maximum power happens in this case when $$R=|Z_t| = 5 \sqrt{5}$$ , and I get that P=6.18, while the answer is P=8.3.
    Where did I went wrong?
     

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  3. Jan 27, 2016 #2

    gneill

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    Staff: Mentor

    There's hard to tell where you went wrong if you don't show the details of your work.

    However, can you show how you found the Thevenin impedance?
     
  4. Jan 27, 2016 #3
    When I look at it again, I get Z_t=5+j5,but I still get wrong solution. Do you want me to post my full (wrong) solution?

    I found 5+j5 (parallel circuit of 10 ohms and j10 ohms)
     
  5. Jan 27, 2016 #4

    gneill

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    Staff: Mentor

    Yes, that would be best.
     
  6. Jan 27, 2016 #5
    So I found thevenin impedance as parallel circuit of 10 ohms and j10 ohms (I can ignore 10 ohms and -j10 ohms since there is voltage source parallel to them).
    Point A is to the left of our resistor , and point B to the right.
    Bottom line of a circuit has potential equal to 0 (i don't know how to say it in english, hope you understand).
    point A has potential 10V. When finding potential of point B, i ignore voltage source of 10V( the on the left). The impedance is parallel circuit of 10+j10 and 10-j10 so Z=10 ohms.
    The current is then 1+j. The current going through 10+j10 is exactly 1 A, so the potential of B is j10. Now U_th= 10-j10.
    R should be 5 sqrt(2) , the current in thevenin circuit should be I=U_th/(Z_th+R) = 1.083 (effective value) and power is P=I^2R=5.86
     
  7. Jan 27, 2016 #6

    gneill

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    Staff: Mentor

    Your Thevenin impedance and voltage look good, as do the resistance and the magnitude of the current that you found. The first problem I can see is with the very last bit, the calculation of the power, P = I2R. Check that calculation again. Break it down by calculation step if necessary.
     
  8. Jan 27, 2016 #7
    Thank you, I was putting R=5 this whole time and not 5sqrt(2). Now it's correct.
     
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