# Find the resistance for maximum power transfer

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1. Oct 6, 2014

### MattHorbacz

1. The problem statement, all variables and given/known data

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. Oct 6, 2014

### 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?

3. Oct 6, 2014

### ehild

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
4. Oct 6, 2014

### MattHorbacz

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

5. Oct 6, 2014

### MattHorbacz

yeah i realized that wouldnt work, but a man can dream =]

6. Oct 6, 2014

### ehild

I think the output power is on RL instead of R...

7. Oct 6, 2014

### MattHorbacz

yes, but we are already given R_L...i need to solve for R to maximize the output power on R_L

8. Oct 6, 2014

### MattHorbacz

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!!!!!!!!!