Not dumb questions at all!
I'll add to Darrell's accurate response. The key being "proportional".
Electrical power is _always_ the product of current and voltage (at an
instant, not average).
V=I*R -> I=V/R, P=I*V -> P=(V/R)*V=(V^2)/R
For AC, the voltage and current can be out of phase, and the power
transfered is a function of this phase difference (called "power
factor"). In the case of a 90 degree difference (purely capacitive or
inductive), there is no power transfer.
So if you tune your antenna perfectly (so it is purely resistive), then
v_peak*sin^2(wt)/R will be accurate. The R (radiation resistance) will
be dependent on the characteristics of the antenna (probably most
importantly, length vs wavelength).
--Eric
On 05/04/2017 02:20 PM, Eric Smith wrote:
On Thu, May 4, 2017 at 1:07 PM, Darrell Harmon
<darrell@xxxxxxxxxxxxxxxxxxxxx <mailto:darrell@xxxxxxxxxxxxxxxxxxxxx>>
wrote:
Yes, transmitted power is proportional to voltage squared (antenna
input impedance is constant with input power variation).
The electric and magnetic fields are proportional to voltage and
current.
Thanks for the quick answer!
