Yes, conceptually, the real power is all in R and will be I^2R where I is in RMS. The real power in X is zero. The reverse happens for reactive power. Just working up to it mathematically is...a bit less obvious.Yes, an AC current source at one frequency.
Assume we know the rms current of the source.
Off the top of my head without thinking about it too much, is the power in R just I^2R, while the power in X is zero?
Ideally... But X will always be a portion... The difference ( phase angle ) has to be added.. If you had 4 watts real power and 3 watts imaginary power you get a phase angle of 30 degrees so a apparent power of 5 watts..Off the top of my head without thinking about it too much, is the power in R just I^2R, while the power in X is zero?
The power company delivers a constant voltage, and then bills me for "real" power provided that the power factor is not too far out...Ideally... But X will always be a portion... The difference ( phase angle ) has to be added...
Hi,Suppose you have load R+jX that is fed with a Current Source.
What is the total power delivered by the current source?
What is the power in the R part of the load?
What is the power in X part of the load?
Reactive impedance never dissipates real power, so I don't understand your question.None of the power is dissipated in the reactive part of the antenna's complex feedpoint impedance?
That is what I concluded, hence I had already answered my own question, and you and others confirmed that answer...Reactive impedance never dissipates real power, so I don't understand your question.