Transformers for Maximum Power Transfer

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To achieve maximum power transfer to a 20-ohm load resistor, the turns ratios a1 and a2 must be determined such that the source impedance matches the complex conjugate of the load. The relationship a = n2/n1 is crucial, with Rs equating to RL for optimal power transfer. The source and load must be real resistors, necessitating the cancellation of reactive components from inductors and capacitors. The first turns ratio should be adjusted to eliminate complex impedance, while the second turns ratio must ensure the source impedance matches the load. This approach leads to a solution for the turns ratios required for maximum power transfer.
aznmax218
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Determine the turns ratios a1 and 12 for maximum power transfer to the 20 ohm load resistor. a=n2/n1; Rs=RL for maximum power transfer, i.e. Zsource = the complex conjugate of the load.
2hxlpfq.jpg
i have try to use
Z1=a1^2*Zs and Z2=a2^2*Z1
and i know Z1=-j/(2*pi*f*c)
 
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aznmax218 said:
Determine the turns ratios a1 and 12 for maximum power transfer to the 20 ohm load resistor. a=n2/n1; Rs=RL for maximum power transfer, i.e. Zsource = the complex conjugate of the load.


2hxlpfq.jpg



i have try to use
Z1=a1^2*Zs and Z2=a2^2*Z1
and i know Z1=-j/(2*pi*f*c)

Welcome to the PF.

Wow, on the surface, that looks like a hard problem.

I'd offer a couple of hints, in hopes that they lead you to the answer. First, notice how the source and load are both real resistors (excluding the inductor and cap). So the inductor and cap have to cancel each other out in terms of complex Z, or else you can't match to the 20 Ohm load. What can you do with the first turns ratio to cancel them out?

And second, assuming that you can cancel out the complex impedances with the first turns ratio, what do you now have to do with the 2nd turns ratio to match the source and load impedances?

Show us your work along those lines, to see if it can work that way...
 

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