Network Maximum Power Transfer Calculation

[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)

 

[gneill]

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

I=E(1000/1220)(330.3)

Close. Both the Thevenin and Load resistors are still there. You've only accounted for one of them.

If you consider the voltage divider created by the two EQUAL resistors, what must be the potential across the load resistor R_L?