Big doubt of the day. For maximum power transfer we use a source impedance equal to the load impedance (conjugate for AC, to be exact). We match for reflection too, but that's another story.

Assuming resistive matching half of the power is on the load side, half on the source side and, with a transmission line, assuming it's lossless and matched, nothing on the trasmission line.

However, usually, matching on the source and/or the load to the line is done reactively (transformers, LC, tees, and so on), with narrow band matching being the theoretical optimum (as in, perfect match assuming perfect components).

How is the power balance in that case? Do I simply use phasors so that the inductive and capacitive reactive power just cancel each other (didn't do the math but that should justify it)? In this case (perfect components and lossless line) the only real power loss would be on the real impedance of the source and the rest should be `usefully' available on the real impedance of the load.

Am I missing something?

  • \$\begingroup\$ That looks about right; when XC and XL cancel, what's left is to pay the bill. \$\endgroup\$ Commented Feb 18, 2021 at 10:05

1 Answer 1


Am I missing something?

No, you are not missing anything. If you design an impedance matching network (L-Pad, pi-network etc.) then it is lossless.

How is the power balance in that case?

If you are matching a resistive source to a different resistive load, the powers are still balanced as per this example: -

enter image description here

Image from this calculator.

In the above example you'll see that it calculates the voltage gain as 2.44949 so, if 1 volt is fed at the input, the output will be 2.44949 volts. If you calculated the load power and input power (1 volt into 50 Ω) it equals the output power.

Of course there is still the same power dissipated in the source impedance but it sounds like you realize that.

If your load is complex then make it resistive with the appropriate conjugate component. Ditto the source impedance or, both.

  • \$\begingroup\$ OK, thank you for the response and example. It seems that I didn't forget too much from the uni days \$\endgroup\$ Commented Feb 19, 2021 at 6:55

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