Understanding Impedance Matching for Optimal Energy Transfer

[h0dgey84bc]

Hi,

I'm having a little trouble understanding the concept of impedance matching to maximise energy transfer. I understand that the reactance of a capacitor is X_c= \frac{1}{\omega C } and that it is always 90 degrees lagging of the current in the complex plane which leads to a capacitors complex impedance being defined as X_c= \frac{-j}{\omega C }. I also understand the reactance of an inductor is X_L= \omega L and since it leads the current by 90 degrees, it has a complex inductance defined as X_L=j \omega L.

Since the reactance of the inductor and capacitor are always antiparallel to each other, but perp to the reactance of a normal resistor, you find the impedence of an LRC circuit to be Z=sqrt( R^2+(\omega L -\frac{1}{\omega C})^2 ).

That's the point where my knowledge of AC circuits ends. Why does matching the output impedance of a generator with my load circuit maximise power transfer?
What are the conditions for this matching?

To make the discussion more concrete my motivation is this question http://grephysics.net/ans/8677/64 from an old GRE paper, that I am trying to wrap my head around.

Please help a poor theorist who hasnt seen circuits for many a year out.

Thanks

 

[berkeman]

Since it is a maximization problem, you would use differentiation to derive the conditions for the maximum.

Start with just a resistor voltage divider -- you have an indeal DC voltage source with an output resistance Ro, and a load resistor to ground Rl. Write the equation for the power transferred to Rl in terms of Ro and the voltage source, and use differentiation to derive the value of Ro that gives the maximum power at the load Pl. For resistors, you will find that Rl = Ro gives the maximum power Pl.

Now do it for a general complex reactances Xo and Xl. The equations are a bit more involved, but you showed above that you are familiar with them. The answer you get is almost like Rl = Ro, but with a twist. See if that gets you the answer or not. If not, post your work and we'll see if we can help more.