Consider two parallel wires A and B of equal lengths, carrying in-phase AC’s. Assume that, the frequency is so low relative to the distance R between them, that the phase delay is ignorable and that there is an even distribution of current on every segment of the wires.

I’m trying to calculate, what would be the total impedance (Znet) on one of the wires when their changing magnetic fields are negatively affecting each other (mutual inductance). I tried to break the net current (at time t) into two parts as you can see below:

One part of the net current is the current which would occure normally caused by the input voltage.

Other part is the current in opposite direction, caused by solely the mutual inductance.

enter image description here

And I summed up the two to find total current. I don’t know if this is a right thing to do. Summing up opposite currents maybe allright with superposition theorem but physically it doesn’t “feel” right or make sense. That’s why I hesitate.

If the idea is right, then I will to try to adapt it to "two parallel antennas situation" to calculate their effective impedances and match the input impedance to that value.

  • \$\begingroup\$ I think you've fallen down at the first sentence - the current in an antenna wire is different all along its length because that is how dipole or monopole antennas work at their resonance. This makes it difficult to answer because of having to roll-back your misconceptions. Maybe you didn't mean antenna and maybe you were just considering low frequencies whose wavelength was well below the length of the wire? \$\endgroup\$ – Andy aka Feb 9 '17 at 18:23
  • \$\begingroup\$ I needed to emphasize on the calculation for the effect of parasitic inductance so I had not cared enough for the poor description of the other variables. The description was not right for the whole of an antenna because there is a current distribution. \$\endgroup\$ – Xynon Feb 16 '17 at 9:14
  • \$\begingroup\$ And it would be right, as you pointed out that, if the frequency was low enough that the instantaneous current values would not vary much across a wire (this wouldn’t effectively be an antenna anymore with such a large “wavelength/wire length” ratio and high capacitive reactance). And there is also the phase delay and a power factor due to it, which further complicates the question unnecessarily. So, I reformed the question and the title as with two wires as you corrected. \$\endgroup\$ – Xynon Feb 16 '17 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.