Hi electrician:That should be real power.
Do you know the correct answer?
Is 97.6^2*.05 anywhere near the correct result?
It is used in determining the total voltage drop along the wire when the current in the wire is known; it's used for determining the reactive power, which is not dissipative.ok. In that case what is the imaginary part of the impedance used for?
Because when you do that you are including the reactive part of the line impedance in the calculation, and the reactive part doesn't contribute to the dissipative power loss. Also, you need to use the RMS value of the current, which is the magnitude of the current phasor neglecting the phase angle.Why does multiplying everything out like I have done give me the wrong answer?