Maximum Average Power Transfer

For the circuit shown in the picture, find the load impedance \(Z_{L}\) that absorbs the maximum average power. Calculate that maximum average power.

So far, I can get the load impedance correct, but I can't get the maximum average power right. I try to turn the circuits on the right of the load into a Thevenin equivalent circuit, but during the process, the value for the source is messed up, so I try to solve it directly with current source, and I used the current divider concept to calculate for the current goes through the \(Z_{L}\), but the maximum average power is not correct, far off from the answer.

Answer: load impedance: 3.415 - j0.7317\(\Omega\) , maximum average power: 12.861W

Please outline how you got the correct load impedance. The key to getting the max avg power (either the voltage across it or the current through it) is probably there.

Also, show (in some detail) how you are going about getting the max average power, even though you are getting a wrong result. That way we can see what you have done correctly and help you spot and get past where you have gone wrong.

Well, I first cut off the load, and turn off the independent current source, which leaves the rest of the circuits simply equivalent to (8+j6)//5 ohms impedance. According to the Max Average Power Transfer Theorem, the Max Average Power Transfer occurs when ZL is the complex conjugate of the ZTh of the rest of the circuits.

Now (8+j6)//5 = 140/41+j(30/41)

So ZL = 140/41-j(30/41)

That's how I obtain the load impedance required for the max average power to occurs.

As for the calculation the maximum average power, I use a few approach.

Norton method:

Because the definition for Norton equivalent circuit is the current flow through a short circuit. So I replace the load with a short circuits, which let the 5-ohm resistor useless in the circuit. So the current flow through the load should be:

6*(j10/((8-j4)//(j10))) = 3+6i

According to the formula, the maximum power transfer should be 0.5*|I|^2*RL(Re(ZL)).

|I|=6.7082, |I|^2=44.9999
0.5*|I|^2*RL=76.8374 W

Thevenin method:

The Thevenin equivalent circuit has an independence voltage source with the value of -2.625-j3.996 = VTh, |VTh|=4.7814
|VTh|^2=22.86178596
Maximum average power transfer = 0.8368 W

These are the method I used.

rogerloh4.0 said:

Norton method:

Because the definition for Norton equivalent circuit is the current flow through a short circuit. So I replace the load with a short circuits, which let the 5-ohm resistor useless in the circuit.

Click to expand...

Your reasoning is just fine, up to this point.

So the current flow through the load should be:

6*(j10/((8-j4)//(j10))) = 3+6i

Click to expand...

Where did this equation come from?

First, I highly recommend that you track your units through your work.

Second, ask if the answer makes sense. What would you expect the current in the load to be if the impedance through the other path was very, very low? For simplicity sake, say it were changed from (8-j4)Ω to j0.01Ω. Would you expect the load currrent to be large, or small? What does your equation result in?

Step back and take it one step at a time:

Q1) With the load shorted, what is the impedance seen by the current source? It's fine to express it as a parallel combination.

Q2) What is an expression for the voltage across the current source?

Q3) What is an expression for the current through the load?

Q4) How does this expression compare to yours?

(And track your units through your work or someone else can help you.)

Q2) What is an expression for the voltage across the current source?

Click to expand...

With RL shorted or opened?

mlog said:

With RL shorted or opened?

Click to expand...

Shorted. These are a progression of questions related to the OPs use of short-circuit analysis to determine the Norton equivalent circuit.

WBahn said:

Shorted. These are a progression of questions related to the OPs use of short-circuit analysis to determine the Norton equivalent circuit.

Click to expand...

I thought mine was a reasonable question, because I don't see the relevance of finding the voltage across the current source. On the other hand, it you want to find the open circuit voltage at the opened load, then I can see the relevance of that. The Thevenin or Norton impedance would be Voc/Isc. Of course it is useful to employ a shortcut to find the impedance from the opened load looking inwards towards the network (with the current source opened).

mlog said:

I thought mine was a reasonable question, because I don't see the relevance of finding the voltage across the current source. On the other hand, it you want to find the open circuit voltage at the opened load, then I can see the relevance of that. The Thevenin or Norton impedance would be Voc/Isc. Of course it is useful to employ a shortcut to find the impedance from the opened load looking inwards towards the network (with the current source opened).

Click to expand...

Oh, it was a quite reasonable question since it wasn't explicitly stated.

The relevance is that the OP appears to have attempted to employ the current divider equation but didn't do it correctly. So the idea was to walk him through the fundamental analysis that results in the current divider equation. Namely, find the total impedance of all the parallel branches together, find the voltage across them, find the current in the one in question.

Thanks for all your helps! I finally solved it!