In the tuning of ~ 25 mh/ 2.6 ohms resonances @ 60 hz we find that it has an
effective impedance for practical considerations Of ~ 10 ohms. This in fact is
the actual impedance to be tuned to capacitively, so we set X(C) to equal 10
ohms.
X(C) = 1/{2pi*F*C}
X(L)= 2pi*f*L
are the equations involved for equal matches at resonance.
Setting X(C) = 10 we find that this value enables a one amp conduction given a
10 volt source.
10= 1/[6.28*60*C]
1/C= 376.8*10= 3768
C= 1/3768 F or 265 uf or microfarad = 2.65 * 10^-6 F.
The value of stored energy in 265 uf is given by that one amp conduction
whereby J = .5CV^2.
Here that one amp conduction is made possible by a 10 volt source placed across
a 10 ohm load.
Ordinarily giving the V value as 10 volts; V^2 becomes 100, but recognizing
that V(peak)^2 = 14.4 squared this instead becomes 207.4, instead of just 100.
J= .5 C V^2= .5* {2.65* 10^-6}*207.4 = .027 joules
This amount of energy is transferred 120 times per second between L and C for
an energy transfer of ~ 3.3 joules/ second.
But ten watts is supposed when ten volts makes a one amp current.
If the inductor had a Q of pi; pi times more current or voltage could ensue;
then the figures would be in agreement. However resistive inductors at certain
frequencies can also have a Q better then pi, in which case the energy
transfers between L and C can exceed the I^2R losses made on the inductors
resistance itself.
This is somewhat a complicated matter whereby perhaps others can explain the
discrepancy better here. Perhaps I have made errors in my mathematics. But this
was the point I was trying to drive across. Thanx for the reply.
HDN
