# What is the impedance of a multi-tap impedance matching transformer when all taps are used at the same time?

So if I have an impedance matching transformer with two taps, one 8ohm and the other 4ohm, and both taps have a winding relationship so that the other side "sees" a 4kohm load, what is the impedance if I use both taps at the same time assuming both loads are correct to their respective taps? what about the power if a signal is applied on the 4kohm side? will it split through each one of the other two loads? with what proportion? • This has the feel of a "homework question", although I suspect it might not be. But in any case, what have you worked out so far? Suppose you connect a fixed AC voltage source to the primary. How much current flows in it when you connect one load at a time? How much current flows when you connect both loads? What does this tell you about the effective impedance of the primary when both loads are connected? (Keep in mind that the turns ratio is the square root of the impedance ratio.) Feb 21, 2018 at 2:27
• I have solved it :), I ran a simulation on proteus and the current through the 4k resistor doubled when both loads were connected which means the resistance halved. The power remained the same across each load. However, for the second case I'm getting some strange readings for the current through the primary, it is much higher than it should be, but it’s probably some glitch in the program since it stays like that with loads connected or not.
– Raz
Feb 21, 2018 at 3:51
• The power readings of each load are the same as for the multi-tap transformer, power stays the same on the secondaries, so power must double across the primary and resistance should be half.
– Raz
Feb 21, 2018 at 3:51
• Yes. Keep in mind that the primary current includes the out-of-phase "magentizing current", which is determined only by the inductance that you selected for the primary and the frequency at which you're driving it. Feb 21, 2018 at 4:05

To produce the same power in a 4 ohm load as an 8 ohm load requires that the 8 ohm load receives an RMS voltage (or DC voltage for resistors) that is $\sqrt2$ as high as the 4 ohm load.