Network Maximum Power Transfer Calculation

[shaltera]

Homework Statement

Determine, for the network shown in fig 4 the value of the load
Impedance that will dissipate the maximum power and the value of
this power.

Homework Equations

Using Thevenin theorem to simplify the circuit then,I believe maximum power transfer theorems apply.

P=I²R
R=√(R²+x²)
I=E/Zt
Zt=z+R

The Attempt at a Solution

Homework Statement

But E is in Hz not in Volts?Do you think it could be a printing error?

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi shaltera! welcome to pf!

But E is in Hz not in Volts?Do you think it could be a printing error?

no

the impedance of a circuit (or of a component) does not depend on the voltage (the emf)

but it does depend on the frequency, ω, doesn't it?

 

[shaltera]

Thevenin Impedance

Thank you for your reply.I calculated Thevenin impedance.Calculation below:

R₁=220Ω
R₂=1KΩ
R₃=150Ω

(R₁xR₂)/(R₁+R₂)=(220ΩX1000Ω)/(220Ω+1000Ω)=180.32Ω

Resistance of L₁
Z_l=2πfL=2xπ50x700x10^-3=220Ω (round)

R₃+RL₁=150Ω+220Ω=370Ω

Z_TH=180Ω+370Ω=550Ω

And then?

 

[gneill]

Careful with the inductor. The inductor has impedance, an imaginary value, not a real values resistance. Z_L = jωL.

 

[shaltera]

How ω can be calculated?There is no ω?

 

[gneill]

How ω can be calculated?There is no ω?

ω is the angular frequency. $\omega = 2\pi f$.

 

[shaltera]

Thank you

 

[shaltera]

ZL = jωL=j2πfL=j220
R3+RL1=150+j220
ZTH=180+150+j220=330+j220

 

[shaltera]

OK.I calculated Zth what should I do next?Thank you

 

[gneill]

ZL = jωL=j2πfL=j220
R3+RL1=150+j220
ZTH=180+150+j220=330+j220

Looks good. Don't forget that the units of the impedance is still Ohms, just like resistance.

 

[shaltera]

Sorry I forgot to add Ω at the end

330+j220Ω

 

[gneill]

Okay! So now you either apply the Maximum Power Transfer Theorem, or you try to derive it from scratch. It's simple for real-valued resistances, but it gets a bit "mathy" for complex impedance values. The result is very simple, but getting there is tedious! You might want to do a little investigation on the web to see how its done.

 

[shaltera]

You mean:

Z_L=Z_TH
P_max=|V_TH|²/8R_TH

Thanks

 

[gneill]

Nope Z_L is not Z_th for maximum power transfer! That's the tricky bit. It's very closely (amazingly, really) related to Z_th though...

Do a web search on "Maximum Power Transfer Theorem".

 

[shaltera]

Thanks again.I keep looking and looking.All examples have a voltage source.

 

[gneill]

Thanks again.I keep looking and looking.All examples have a voltage source.

A voltage source and a series resistance or impedance. In other words, a Thevenin model...

Check out the Wikipedia entry on the MPTT. It has a section devoted to impedance.

 

[shaltera]

Searched in Wiki on the MPTT,weird results.I'm fed up now,spent too much time on this silly problem.Thanks anyway

 

[gneill]

Searched in Wiki on the MPTT,weird results.I'm fed up now,spent too much time on this silly problem.Thanks anyway

Did you come across this URL?

http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

 

[shaltera]

Yes I did, but examples are with a voltage source.I'm sure you know what you are talking about,its probably me.I really wanted to do it myself but its getting ridiculous and I have to move to Plan B.

 

[shaltera]

I started college 4 weeks ago, and they want me to solve this problem,ridiculous. That's why people noways do not remember simple things.Colleges,Uni give so much information for short time.Electrical and Electronic Theory in some countries is divided in 3 part (12 months), in UK in 7 weeks.

 

[gneill]

Yes I did, but examples are with a voltage source.I'm sure you know what you are talking about,its probably me.I really wanted to do it myself but its getting ridiculous and I have to move to Plan B.

Your circuit also has a voltage source. It's an AC voltage source, only they haven't specified a particular voltage; they've just supplied the frequency of the AC source. The frequency is the important piece of information for this problem since it determines the impedance of any reactive components (inductors or capacitors). You can assume a 1V AC source if you wish.

While the total power depends on the voltage, the fraction of the total that ends up in the load depends upon the source impedance (your Thevenin equivalent) and the load impedance. The source voltage is just a scaling factor. The frequency determines the impedances, and so determines how the power will be split between the source impedance and load impedance.

 

[shaltera]

what a headache?

I can't find a solution

 

[gneill]

If the problem is just looking for the required load impedance then you are free to use the result of the Maximum Power Transfer Theorem and be done with it. Just state that that is what you used; it's a well known theorem. It's like using KVL or KCL or Ohm's law, which you don't derive every time you use them.

If you are required to derive the MPTT from scratch, that is another matter.

 

[shaltera]

The maximum power transfer theorem states: A load will receive maximum power from a network when its resistance is exactly equal to the Thevenin resistance of the network applied to the load.

 

[gneill]

Yes, that's true for resistances. There's a bit of a twist when impedances are involved since there are separate real and imaginary quantities that interact through the mathematical operations. The result is still very simple though. Look at the last line of the Wikipedia article...

 

[shaltera]

I've looked at it,Zl=Zs.Too complicated

 

[gneill]

I've looked at it,Zl=Zs.Too complicated

Not Zs. Zs*. The "*" designates the complex conjugate (which you've used before, I'm sure, when you want to clear complex values from denominators).

Zs is the source impedance: your Thevenin impedance.

 

[shaltera]

I think is something very easy, but I don't get it.

 

[gneill]

I think is something very easy, but I don't get it.

Is there something in particular that stumping you?

The setup in the Wiki article is identical to that of your simplified circuit: a voltage source with a series impedance driving a load impedance. The only difference is the variable names assigned.

 

[shaltera]

Just don't get.I give up.

 

[shaltera]

Thanks.And sorry for wasting your time

 

[gneill]

Thanks.And sorry for wasting your time

You're welcome. No worries, glad to help (or at least try ) If you want to continue or try again, we'll be here.

 

[shaltera]

Thanks.This problem is already solved..

 

[shaltera]

Your circuit also has a voltage source. It's an AC voltage source, only they haven't specified a particular voltage; they've just supplied the frequency of the AC source. The frequency is the important piece of information for this problem since it determines the impedance of any reactive components (inductors or capacitors). You can assume a 1V AC source if you wish.

While the total power depends on the voltage, the fraction of the total that ends up in the load depends upon the source impedance (your Thevenin equivalent) and the load impedance. The source voltage is just a scaling factor. The frequency determines the impedances, and so determines how the power will be split between the source impedance and load impedance.

What do you me by to "assume a 1V AC source"?

E_th=1V?

 

[gneill]

What do you me by to "assume a 1V AC source"?

E_th=1V?

Sure. 1 V @ 50 Hz would do nicely.

If a particular value is not critical to the details of the analysis, it's generally simplest to assume a unit quantity.

 

[shaltera]

All examples I found on the internet are with a load resistance Rl,resistors and voltage

 

[gneill]

All examples I found on the internet are with a load resistance Rl,resistors and voltage

What sort of examples are you looking for? Worked problems?

 

[shaltera]

Similar circuit.What confuses me is that there is no Voltage source and between A-B is not resistor so I can't calculate Rl.This circuit diagram looks to me like an open circuit.Not to forget this 700mH inductor.

 

[gneill]

Similar circuit.What confuses me is that there is no Voltage source and between A-B is not resistor so I can't calculate Rl.This circuit diagram looks to me like an open circuit.Not to forget this 700mH inductor.

There is a voltages source: A 50 Hz AC voltage source. You can assume any value for the voltage you wish, it will not affect the results.

The inductor just contributes to the impedance attributed to the supply network. Once you reduce the supply network to its Thevenin equivalent as seen from terminals A-B the circuit then looks just like the archetypal "maximum power transfer circuit where you are to stick some impedance on A-B to serve as the load.

 

[shaltera]

I have no problem to understand this circuit diagram, and find the Zl?
But with Zth I don not get it,sorry

 

[gneill]

That's why I directed you to the Wikipedia article. It has a relatively clear derivation of the result where the R's are instead impedances.

One way to grasp the reason for the result found is that reactances do not consume real power. They store and return it (a capacitor charges, storing up energy (1/2)CV² on one half of the cycle, but return that same power to the source on the discharge half of the cycle. The same goes for inductors) The power that accomplishes this, while not lost, is not consumed by the load, and the source still has to be able to provide and recover it on every cycle.

So the way to avoid this waste is to choose the reactive part of the load impedance so that when added to the source impedance the net result is purely resistive. Then you're left with just the source resistance and load resistance to deal with, which is the case solved in your post (I see you've changed that. Oh well...).

 

[gneill]

The power dissipated should end up being a real value only. Imaginary power is not dissipated by loads.

You still have not taken in the conclusion of the derivation shown on the Wikipedia page!

You correctly calculated the source impedance in post #8. The required load impedance can be written by inspection from that value. Look again at the last line of the derivation on the Wikipedia page. It tells you how to write the load impedance from the source impedance; they are complex conjugates.

The answer you supply for the actual power dissipated should be a real value, and it should leave the source voltage in symbolic form ("E").

 

[shaltera]

I= Vs / Zs+Zl
Zl=Rl+jXl
P=I²Rl

Zs=Rs+jXs=Rl-jXl

 

[gneill]

I= Vs / Zs+Zl
Zl=Rl+jXl
P=I²Rl

Zs=Rs+jXs=Rl-jXl

Okay, can you put numbers to the variables that you have values for?

 

[shaltera]

C=1/(\omegaX_c)

 

[shaltera]

I managed to calculate Ze and Zl

Ze=(330.3+j219.8)ohms
Hence Zl=(330.3-j219.8)
C=1(314x219.8)=14.49mF

14.49mF is the power I believe

 

[gneill]

I managed to calculate Ze and Zl

Ze=(330.3+j219.8)ohms
Hence Zl=(330.3-j219.8)
C=1(314x219.8)=14.49mF

14.49mF is the power I believe

mF is capacitance. That's the equivalent capacitance that the load impedance presents. The unit of power is the watt. You'll have to find the power dissipated by the resistive part of the load impedance.

Since you weren't given a particular value for the source voltage your result will be an expression rather than a numeric value. Give the source voltage a name, say E. Now, knowing all the rest of the component values for the circuit you can determine the currents and potentials in terms of E.

Hints:
1. You've determined the Thevenin impedance for the source. Determine also the Thevenin voltage E_th in terms of E.
2. With the Thevenin equivalent in place and the load impedance connected you have a series circuit. Determine the net impedance that the Thevenin source "sees".

 

[shaltera]

Rl=Re[Zth] and Xl=Im[Zth]

Pmax=Vth2/4Re[Zth]

 

[gneill]

Rl=Re[Zth] and Xl=Im[Zth]

Pmax=Vth2/4Re[Zth]

Yes. But you can simplify it further by plugging in the known value of Re(Zth) and an expression for Vth in terms of E. You should end up with an expression of the form c₁E² or E²/c₂, where the c's are some constants.

 

[shaltera]

Zth=330.3+j219.8

Pmax=1V2/4(330.3-J219.8) ?