Network Maximum Power Transfer Calculation

[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?