We have the circuit in the figure.I have to find n for which RN has the maximum power.What is the power?We have that J=2A,R1=1 Ohm,R2=75 Ohm,RN=100 Ohm enter image description here

I find the Thevenin equivalent for the left part of the circuit,where V=J*R1=2V

The reflected load impedance is \$ZL’ = \frac{ZL}{n^2}=\frac{175}{n^2}\$

For maximum power transfer,I have to equal this to ZL' but is ZL'=R1?

After I learn how to find n there,its easy to find the power...but how do I find n?

  • 1
    \$\begingroup\$ Be careful. The question is asking for the maximum power in RN, not the maximum total power in R2+RN. As such, you need to consider R2 as part of the source resistance, not the load resistance! \$\endgroup\$
    – Dave Tweed
    Feb 3, 2015 at 0:06

2 Answers 2


Here's a better drawing of your circuit:


simulate this circuit – Schematic created using CircuitLab

The maximum power transfer to the load occurs when the load impedance matches the source impedance. The load impedance is obvious — 100Ω — but the source impedance is R2 in series with whatever impedance the combination of I1, R1 and XFMR1 represent.

Impedances in series simply add, so the total source impedance connected to the load is the effective impedance at the right side of XFMR1, plus R2. Therefore, we need the impedance of XFMR1 to be RN - R2 = 100Ω - 75Ω = 25Ω.

The actual source impedance of I1 and R1 together is just R1, or 1Ω. Therefore, we need the transformer to convert 1Ω on the left to 25Ω on the right. A transformer converts impedances in proportion to the square of its turns ratio n, so we need to know what value of n will create a 1:25 transformation:

$$n^2 = \frac{25\Omega}{1\Omega}$$

$$n = \sqrt{25} = 5$$

  • \$\begingroup\$ I love you and I dont deserve you. \$\endgroup\$ Feb 3, 2015 at 21:17
  • \$\begingroup\$ Also,how can I find the current that runs through RN? \$\endgroup\$ Feb 3, 2015 at 21:28
  • \$\begingroup\$ If you can't figure that out from here, you're not even trying. Hint: What is the relationship between the voltages on either side of the transformer? \$\endgroup\$
    – Dave Tweed
    Feb 3, 2015 at 22:30
  • \$\begingroup\$ I guess I can find it this way : -V+R1*I +R2'*I + Rn'*I=0 And here I find I? \$\endgroup\$ Feb 6, 2015 at 21:50

I think you are correct in your analysis so far. You already have the equations necessary to find n. \$Z_{L}' = \frac{Z_L}{n^2} = R_1\$. Then solve for n.

EDIT: My original answer above is incorrect. Since the question asks for the power across \$R_n\$ (thanks @DaveTweed) , the key is to split the total output load into two parts, \$R_n\$ and \$R_2\$, combine the reflected impedance of \$R_2\$ with \$R_1\$, and match that to the reflected impedance due to \$R_n\$. That should give you a formula with n on both sides, which you can then solve for.

@DaveTweed already explained his way - I just want to explain how I thought about the problem. The picture below describes how you can transform the load impedance into a reflected impedance across the transformer, depending on its number of turns.

enter image description here

So, do the same thing with your load impedance. I denoted the reflected impedances with primes.


simulate this circuit – Schematic created using CircuitLab

Now the circuit is a simple voltage divider, and you know that to get the maximum power transfer you must match Rn' with R1 + R2'. Therefore:

$$ R_1 + \frac{1}{n^2} R_2 = \frac{1}{n^2} R_n $$ $$ n = \sqrt{\frac{R_n - R_2}{R_1}} $$

The current through \$R_{n}'\$ should be easy to find now.

  • \$\begingroup\$ See my comment above. \$\endgroup\$
    – Dave Tweed
    Feb 3, 2015 at 0:11
  • \$\begingroup\$ Thanks Dave, that was a good catch. I'll post an updated solution soon. \$\endgroup\$ Feb 3, 2015 at 0:27
  • \$\begingroup\$ The formula I get is n^2*R1 = R2 + RN \$\endgroup\$ Feb 3, 2015 at 9:49
  • \$\begingroup\$ @Notyourthing: No, that's incorrect. If I tell you that the sign of R2 is wrong, can you figure out why? \$\endgroup\$
    – Dave Tweed
    Feb 3, 2015 at 11:38
  • \$\begingroup\$ I dont understand why actually :( \$\endgroup\$ Feb 3, 2015 at 12:47

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