# Homework Help: Maximum power transfer

1. Jun 17, 2013

### Color_of_Cyan

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

[Broken]

a) For the circuit shown, determine the impedance ZL that results in maximum average power transfer to the ZL

b) What is the maximum average power transferred to the load impedance?

2. Relevant equations

Thevenin, Norton procedures, voltage division, current division, etc etc

Max. average power = 1/2(Vs/Rs + RL)2 * RL

PL = (V/(Rs + RL))2*RL

Rs = RL

Js = -JL

3. The attempt at a solution

I actually do not know how specifically for here how to get the equivalent impedance for the load, but I assume it's the same as finding Thevenin resistance so...

5Ω || 20Ω = 4Ω

4+3j || -6j =

=(1/4+3j + -1/6j)-1

=[ (6j-4-3j)/(24j-18) ]-1

= 24j-18 / 3j-4

multiplying by conjugate now:

=> * (-3j-4)/(-3j-4)

= (-72j2 - 96j + 54j + 72)/(-9j2 - 12j + 12j + 16)

= (72 + 72 -42j)/25

= 144-42j/25

=5.76 - 1.68j

RTh = 5.76Ω - 1.68jΩ

ZL = 5.76 + 1.68jΩ

Now

average power = 1/2*(20/(5.76 + 5.76))2*5.76 = 8.68W

Not sure if this is right

Last edited by a moderator: May 6, 2017
2. Jun 18, 2013

### Staff: Mentor

You didn't complete the Thevenin model; the voltage driving the load impedance through the Thevenin impedance will not be the original Vs. I don't get the result you did.

If it were me, rather than memorize another special case formula to determine the power delivered, I'd use the Thevenin equivalent and the determined load impedance to find the current I through the load, then find the complex power delivered as $P = I \cdot I^*\cdot Z_L$, where I* is the complex conjugate of I. The average power is the real component of the complex power.

3. Jun 19, 2013

### Color_of_Cyan

Ok, my bad I flipped out there. When finding VTh you have to forget about RTh, and find parallel impedances all over again.

The load impedance is still the same as RTh though right? With the imaginary component negated?

So for VTh now, the load cut out will put -j6Ω + j3Ω = -j3Ω

Now the circuit's impedance (just for finding VTh would look like

( -j3Ω || 20Ω) + 5Ω

So ( -j3Ω || 20Ω)

= (1/20Ω - 1/3Ωj)-1

= [ (3Ωj - 20Ω)/(60Ω2j) ]-1

= (60Ω2j)/(3Ωj - 20Ω)

Multiplying by complex conjugate now

*(-3Ωj - 20Ω)/(-3Ωj - 20Ω)

= (-180Ω3j2 - 1200jΩ3)/(-9Ω2j2 + 400Ω2)

= (180Ω3 - 1200jΩ3)/(9Ω2 + 400Ω2)

= (180Ω3 - 1200jΩ3)/(409Ω2)

= 0.44Ω - 2.93Ωj

Now (0.44Ω - 2.93Ωj) is in series with 5Ω

So with voltage division:

VTh = 20∠0 *(0.44 - 2.93j)/(5.44 - 2.93j)

In polar form:

VTh = 20∠0 *(2.96∠-81.4)/(6.18∠-28.3)

VTh = 9.57∠-53.1 deg

So with this and RTh connected to the load, would the current then be I=V/R, and assuming the load is almost the same as Rth with the imaginary component negated and them both combined now:

= (9.57∠-53.1)/(5.76Ω + 1.68Ωj + 5.76Ω - 1.68Ωj),

(Which would then be)

= (9.57∠-53.1)/(11.76∠0)

= (0.81∠-53.1)A ?

How would the complex conjugate be applied to this now?

Last edited: Jun 19, 2013
4. Jun 19, 2013

### Staff: Mentor

Your approach to finding the Thevenin voltage doesn't make sense to me. You want to find the open circuit potential at terminals a-b, yet your first step seems to "disappear" terminals a and b by incorporating the capacitor's impedance with the inductor's.

One approach to finding the Thevenin equivalent for this sort of layout, which I find useful, is to work from the source forward to the open terminals incorporating the components successively into a Thevenin model. So begin with just the voltage divider presented by the source and two resistors.

Here's a diagram where the first step has been completed:

#### Attached Files:

• ###### Fig1.gif
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5. Jun 19, 2013

### Color_of_Cyan

Now I'm kind of confused. I thought you were supposed to disappear / cut out the load to find VTh? I guess I still have trouble understanding it (or for whatever this case is).

Or maybe I just took the voltage divider for the wrong thing?

The voltage division above was for the j3, 20, and -j6 simplified together (with the load cut out).

I take it though that a-b only just meant across the -j6 then?

If so (and assuming the work itself still isn't wrong) I think I could still find it since I think I still found the Vin for all of that to be

9.57∠-53.1 deg

with the voltage divider

so then I think

= 9.57∠-53.1°*(-6j / -6j + 3j)

= 9.57∠-53.1°*(-6j/-3j)

= 9.57∠-53.1°*(6∠-90° /3∠-90°)

=9.57∠-53.1°*(2∠0°)

=(19.14∠-53.1°)V ? Or still all wrong?

Last edited: Jun 19, 2013
6. Jun 19, 2013

### Staff: Mentor

Yes, remove the load but don't "lose" the connection points for the load -- you want to find the voltage at those open connection points.
Could be.
Yes.
Still doesn't look right.

Let's try going step by step in the manner that I proposed.
What do you get for E1 and Z1 considering only the original source and the resistor divider? (looking towards the source from the point of view (1) ).

7. Jun 19, 2013

### Color_of_Cyan

Not sure what you mean, but I guess you simplify out the 20Ω somehow.

Can you switch the left there to a Norton equivalent and then back again to a Thevenin equivalent with a different voltage than before?

8. Jun 19, 2013

### Staff: Mentor

Norton and Thevenin models are always interchangeable. But really, why would you go to the trouble of finding the Norton equivalent first when you've got such a simple voltage divider?

What's the Thevenin equivalent of the above?

#### Attached Files:

• ###### Fig1.gif
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9. Jun 19, 2013

### The Electrician

So far, so good.

10. Jun 19, 2013

### Color_of_Cyan

Okay I see what you mean now; when finding VTh you can break them all up in to other Thevenin models.

RTh of only the 20 and 5 with the source is

(1/5 + 1/20)-1 = 4Ω

and then the VTh of that is

20*(20/25) = (16∠0°)V ?

Would the voltage division for the actual VTh of the problem then be

= (16∠0°)V*(-6jΩ/(4Ω + 3jΩ - 6jΩ) ?

= (16∠0°)V*(-6jΩ/(4Ω - 3jΩ)

= (16∠0°)V*(6∠-90°/5∠-36.8°)

= (19.2 ∠-53.2°)V ? Almost the same as I had originally.

11. Jun 19, 2013

### Staff: Mentor

Yup. That's the idea.
Yes.
Yup. That's right.

12. Jun 20, 2013

### Color_of_Cyan

So

RTh = 5.76Ω - 1.68jΩ

so ZL = 5.76Ω + 1.68jΩ

VTh = (19.2 ∠-53.2°)V

Are all of the above connected in series now?

If so the current I will be:

=(19.2 ∠-53.2°)V/(5.76Ω - 1.68jΩ + 5.76Ω + 1.68jΩ)

=(19.2 ∠-53.2°)V/(11.52Ω)

=(19.2 ∠-53.2°)V/(11.52Ω ∠0°)

I = (1.69∠-53.2°)A

Would I have to convert the current back to rectangular form? Or is there a way to bypass that and multiply I with I* in polar form? Would the conjugate of the current just be (1.69∠53.2)A?

13. Jun 20, 2013

### Staff: Mentor

Yes they are. They form a classic voltage divider with an ideal source, a source impedance, and a load impedance.
Okay, that looks pretty good. You might want to remember to hang onto more digits of precision through intermediate steps (round only results for presentation). This is especially true when there are a lot of steps involved, like switching back and forth from rectangular to polar form. The third digit for the magnitude and angle are both off a tad.
Sure. Think about what happens to a phasor on the complex plane when you form the conjugate; since only the imaginary coordinate is negated the phasor is reflected about the real axis. This is
exactly the same as negating the angle in polar form.

14. Jun 20, 2013

### Color_of_Cyan

Okay so

P = (1.69∠-53.2°)A*(1.69∠53.2)A*(5.76Ω + 1.68jΩ)

If I'm just finding average power though can you just say

Pavg = (1.69∠-53.2°)A*(1.69∠53.2)A*(5.76Ω), because the 5.76Ω is the real resistance?

So Pavg = 19.4W (or approximately) ?

15. Jun 20, 2013

### Staff: Mentor

Okay.
The imaginary parts of all the expression "interact" with the real ones, so you should carry out the full calculation (convert the load impedance to polar form for convenience). The average power will then be the magnitude of the result multiplied by the cosine of the result's angle (that is, the real component of the complex power). Or, if you do the calculation in rectangular form, just pick out the real part of the result for the real (average) power delivered.

16. Jun 20, 2013

### Color_of_Cyan

Ah, thank you so much for that.

= (1.69∠-53.2°)A*(1.69∠53.2)A*(5.76Ω + 1.68jΩ)

= (1.69∠-53.2°)A*(1.69∠53.2°)A*(6∠16.2°)Ω

P = 17.1366∠16.2°

r = (mag)cosθ

Pavg = 17.1366cos16.2

Pavg = 16.45W

Thanks again

17. Jun 20, 2013

### The Electrician

The correct answer is exactly 16 watts.

18. Jun 22, 2013

### Color_of_Cyan

Is there any chance I can know where the error in significant figures is?

I just did the whole problem over again and got the exact same answer, the only thing I got different was rounding off the angle for the current, so it was 53.13 instead of 53.2 but I would just cancel that out.

19. Jun 22, 2013

### The Electrician

In post #12, you have:

"If so the current I will be:

=(19.2 ∠-53.2°)V/(5.76Ω - 1.68jΩ + 5.76Ω + 1.68jΩ)

=(19.2 ∠-53.2°)V/(11.52Ω)

=(19.2 ∠-53.2°)V/(11.52Ω ∠0°)

I = (1.69∠-53.2°)A"

That final result should be I = (1.66667∠-53.13010°)A because 19.2/11.52 is 1.66667, not 1.69, and the angle is slightly different.

Then in post #16, you have:

"Ah, thank you so much for that.

= (1.69∠-53.2°)A*(1.69∠53.2)A*(5.76Ω + 1.68jΩ)

= (1.69∠-53.2°)A*(1.69∠53.2°)A*(6∠16.2°)Ω

P = 17.1366∠16.2°

r = (mag)cosθ

Pavg = 17.1366cos16.2

Pavg = 16.45W"

This should be:

Ah, thank you so much for that.

=(1.66667∠-53.13010°)A*(1.66667∠53.13010°)A*(5.76Ω + 1.68jΩ)

= (1.66667∠-53.13010°)A*(1.66667∠53.13010°)A*(6∠16.26020°)Ω

P = 16.66667∠16.26020°

r = (mag)cosθ

Pavg = 16.66667cos16.26020°

Pavg = 16.00W

20. Jun 22, 2013

### Color_of_Cyan

Thanks again. That's not rounding error but dumb mistakes I keep on making ;/