If the voltage on the output of the coil is higher than that on the input

then

(for constant power) the current must be less.

Nick
G4WHO

Hi Nick,

That would be through for DC, where P = U x I
For AC however it is P = U x I x cos(A), where A is the phase angle between
U and I.

Eg. :
Assume you have an antenna with a capacitance of 300pF and a system loss of
50 Ohm. To bring the 300pF to resonance at 137kHz by putting a 4.5mH
loadingcoil in series. The reactance of the loadingcoil will be about 3.9
kOhm.
Next assume you run the antenna with a 200W TX, so the TX output voltage (=
voltage at the 'cold end' of the loading coil) will be 100V and the current
at the 'cold end' of the coil will be 2A (U and I are in phase).
Further assume that you have built a 'perfect' (lossless and non-radiating)
coil, then the current at the 'hot end' of the coil would still be 2A
(where would any current 'escape' ?) and the voltage would be about 7.9kV
(7.8kV built up over the coil + 0.1kV at the 'cold end'.
Since you are right with you constant power statement (at least for or
perfect coil), the power at the 'hot end' is still 200W. In order to have a
power of 200W with 7.9kV and 2A the phaseshift between U and I must be 89.3
degrees.

73, Rik