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Find the resistance for maximum power transfer

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Steif.ch03.p74.jpg
    Find the value of R where RL equals 50 Ω, for maximum power transfer
    I= 2A, R_L=50 ohms
    2. Relevant equations
    R_th=V_oc/I_sc
    KCL,KVL,V=IR

    3. The attempt at a solution
    First, I created an open circuit including everything except R_L. Then i tried to solve for the Open Circuit Voltage,V_oc. Since current is only flowing through the left loop, it is safe to assume that I_x = the given current which is 2A. Therefore the voltage source =10V. The voltage on the bottom of the circuit will be 0, since it is grounded. I think the top voltage should be v1+10V. V1 would be i_x*40 which is 80. It is going in the opposite direction, so -80. That gives me V_oc=-70.
    Next I closed the circuit, still excluding R_L, and solved for the short circuit current with mesh analysis. I called the left loop i1 and the right loop i_sc. So it follows that
    i1=2
    i_sc-i1=i_x
    i_sc*(40+R)-i1*40=5*i_x
    solving for i_sc gives
    i_sc=90/(45+r)
    My teacher told us that the power transferred is at a max when R_thevenin=R_Load, so
    R_thevenin=-70/[90/(45+R)]=50
    Solving for R I get R=-51.4 ohms, which is wrong...what am i doing wrong? Thank you!
     
  2. jcsd
  3. Oct 6, 2014 #2

    gneill

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    Staff: Mentor

    Your problem occurred when you stated "V1 would be i_x*40 which is 80. It is going in the opposite direction, so -80." What did you mean by it going in the opposite direction? Opposite direction to what?
     
  4. Oct 6, 2014 #3

    ehild

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    Homework Helper
    Gold Member

    Better to solve the circuit with mesh analysis, and derive the expression for the output power and the input power as function of R.

    ehild
     
    Last edited: Oct 6, 2014
  5. Oct 6, 2014 #4
    I thought that the current i_x was directed toward the ground, whereas the voltage i was looking for was at the top...so i guess that is were i got that from...i was reworking it, and i got i_x to be +80, which makes the problem have a nicer solution...didnt try it yet tho
     
  6. Oct 6, 2014 #5
    yeah i realized that wouldnt work, but a man can dream =]
     
  7. Oct 6, 2014 #6

    ehild

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    I think the output power is on RL instead of R...
     
  8. Oct 6, 2014 #7
    yes, but we are already given R_L...i need to solve for R to maximize the output power on R_L
     
  9. Oct 6, 2014 #8
    OMG I GOT IT RIGHT! THE ANSWER WAS 5 OHMS....gneill was right, the problem was in the negative sign. Thanks! you guys are the best!!!!!!!!!
     
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