Network Maximum Power Transfer Calculation

[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/($\omega$X_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) ?

 

[gneill]

Zth=330.3+j219.8

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

There should be no imaginary values in the consumed power. Consumed power is a real quantity. What did you get for your Thevenin voltage? (it should be an expression containing the variable E, which is the unknown source voltage).

 

[shaltera]

e=(V1xZ2)/Z1+Z2

But I don't have a source voltage, just a frequency of 50Hz which I used to calculate Xl?

OK if I assume that my source has avoltage of 1V,then e=1000/1220=1.2V (th voltage)

Pmax=1.2²/(330.3+j219.8)

 

[gneill]

e=(V1xZ2)/Z1+Z2

But I don't have a source voltage, just a frequency of 50Hz which I used to calculate Xl?

OK if I assume that my source has avoltage of 1V,then e=1000/1220=1.2V (th voltage)

Pmax=1.2²/(330.3+j219.8)

Just let the source voltage be the symbol "E". Then your Thevenin voltage is (1000/1220)E.

Now, what impedance does this Thevenin voltage "see" when the circuit is complete?

 

[shaltera]

I'm not sure.Is it Z2?

 

[gneill]

Draw the circuit --- components are Vth, Zth, ZL. Plug in the values of Zth and ZL. What is Vth "seeing"?

 

[gneill]

What's the net equivalent impedance?

 

[shaltera]

Zth?

 

[gneill]

Zth?

What is Zth? What is the load impedance ZL? How are they connected in the circuit? What then is the net impedance?

 

[shaltera]

Zeq=Zl(Vl/Vs-1)

 

[gneill]

Zeq=Zl(Vl/Vs-1)

There are no voltages in an impedance. You have two impedances: Zth and ZL. What are they (you spent much time calculating them!)?

 

[shaltera]

Zth=(330.3+j219)
Zl=(330.3-j219)

 

[gneill]

Zth=(330.3+j219)
Zl=(330.3-j219)

Okay, and how are they connected in the circuit?

 

[shaltera]

In series

 

[shaltera]

Znet=Zth+Zl=660.6?

 

[gneill]

Znet=Zth+Zl=660.6?

Yes!

Note that the reactive components have completely disappeared. The choice for ZL as the complex conjugate of Zth has ensured that the reactive components cancel each other.

So for purposes of analysis at this point, the source impedance is just a real resistance of 330Ω, and the load impedance is just a real resistance 0f 330Ω. Two resistors! Easy to analyze!

Can you find the power in that load resistor? Remember that the source voltage is Vth.

 

[shaltera]

P=i²Rl

 

[gneill]

That'll work; So how will you calculate the current? Remember, the load circuit now comprises the Thevenin voltage (1000/1220)E and two resistors of the same size (330 Ω).

(Actually, the notion that the two resistors form a voltage divider might give you another path to finding the power, particularly since it's a very simple voltage divider with equal value resistors...)

 

[shaltera]

I=Vth/(Zth+Rl)

 

[gneill]

I=Vth/(Zth+Rl)

Remember: The voltage source Vth sees ONLY the resistances. The reactive (imaginary) part of Zth has "disappeared", cancelling with the reactive part of the load impedance.

 

[shaltera]

Zth=(330.3+j219)
Zl=(330.3-j219)

I=E(1000/1220)(330.3)