Re: LF: Lost current in a coil

[Dick Rollema <dickrollema@casema.nl>]

To All from PA0SE

Rik is right of course.  I had overlooked the fact that the voltage
over the coil is almost 90 degrees out of phase with the voltage out of
the transmitter.  So the two voltages cannot be simply added. It
must be done as vectors: SQRT(70^2+8541^2) = 8451,3 V.

73, Dick, PA0SE


At 17:04 2-3-05, you wrote:
> Hello Dick, Mike
>
> it is indeed a "phase thing".
> If a 100 W TX causes a 2 A current in a 5mH coil (XL = 4270 Ohm) and
> thus causes a voltage of 8540 V over the coil then at the "hot"
> end 
> the phase difference between U and I is 89.7 degrees :
> P = U * I * cos(x) thus x = arccos(P/(U*I))
> (assuming no coil losses)
> At the "cold" end voltage and current are in phase.
>
> 73, Rik
>
> > To Mike and All
> > 
> > At 15:15 2-3-05, G3XDV wrote:
> > >A useful test to see whether radiation causes the loss, would be
> to
> > >completely screen the loading coil.
> > Because the size of the coil is so small, expressed in 
> > wavelength,  radiation is negligible.
> > 
> > >I have never understood why, if the current is thought to stay
> the
> > >same from top to bottom of a coil, the voltage is much bigger
> at
> > the
> > >top? Is this a phase thing?
> > 
> > The voltage at the bottom is equal to the output voltage of the
> > transmitter; at e.g. 100W this is 70V over 50 ohm.
> > 
> > As an example let the inductance of the coil  be 5mH. Then
> its
> > reactance at 
> > 136kHz is 4270 ohm.
> > With an aerial current of for instance 2A flowing through the
> coil
> > the 
> > voltage at the top of the coil will be
> > 2 * 4270 = 8541V higher than at the bottom.
> > So total voltage at the top will be 70 + 8541 = 8610V.
> > 
> > 73, Dick, PA0SE
> >