I've been deriving transmission line theory from scratch, out of curiosity, and I'm a bit hung up on one seemingly insignificant point.

When determining the real power drop on a reactive load with a reactive transmission line attached (and supplied by a voltage source), it is commonly accepted that the current amplitude in the loop should be multiplied by only the real part of the load impedance. This does not make sense to me. I think the current amplitude in the loop should be multiplied by the magnitude of the load impedance (r^2 + x^2)^(0.5).

What the hell is going on here!?

  • 1
    \$\begingroup\$ Note that an impedance has a magnitude even if it has no real part, and without any real parts, a circuit is lossless. E.g. lossless transmission line out to infinity. \$\endgroup\$
    – Kaz
    May 12, 2013 at 0:27
  • \$\begingroup\$ Current x impedance gives voltage across the load but it doesn't tell you the power taken by the load. If the load is X + R, the power taken by the load is I^2.R - it has nothing to do with X. \$\endgroup\$
    – Andy aka
    May 12, 2013 at 10:13

1 Answer 1


This does not make sense to me

Go back to the time domain and see that, for the real part of the load impedance, power is always absorbed. The average power is, thus, non-zero.

However, for the imaginary (reactive) part of the impedance, power is alternately absorbed and then delivered (the power alternates from positive to negative). For the reactive part, the average power is, thus, zero.

  • \$\begingroup\$ +1. For the imaginary part, I would say stored and then delivered rather than absorbed to make it clear that there is no energy lost from the electrical system. Both inductors and capacitors can store energy - I hope this clarifies. \$\endgroup\$
    – user16324
    May 12, 2013 at 13:04
  • \$\begingroup\$ @Alfred Centauri This is a great response. Thank you for your time. \$\endgroup\$
    – nick_name
    May 12, 2013 at 14:11

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