To browser these website, it's necessary to store cookies on your computer.
The cookies contain no personal information, they are required for program control.
  the storage of cookies while browsing this website, on Login and Register.

Storing Cookies (See : ) help us to bring you our services at . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: overunity in resonance ?  (Read 5051 times)

Offline hartiberlin

  • Administrator
  • Hero Member
  • *****
  • Posts: 8018
    • free energy research
overunity in resonance ?
« on: August 22, 2005, 07:32:05 PM »
I got the following question:

Thank  you  for your  responce  stephan.
      Let  input voltage   v  = V sin(wt)   ;
                      effective  resistance  seen  by the  driver circuit   
                                    =  resistance  in the primary ( assumed  to  be  very small)
                                           + load resistance in the secondary
                                    =    r  +  R     =   R'  (say)

   now  at  resonance   (after  tunning  the  whole  circuit ),
          current   in the  pimry circuit   i   ~   v/R'
  the  voltage  across  the  inductor   =    quality  factor  times the  inputvoltage   

                                                     =   Q * v

    Therfore   simalar  voltage  Q*v   and current  I  appears  across  the  secondary  coil  of  1:1  transformer    as well as   across  load  resistance   R .
        Now ,
    power  supplied  by the  source  P =  ( V * I ) / 2     (product   r.m.s  values )
         ( also  it  is  known that  at  resonance  power  factor = 1   (approximatly)  )

  power  received  by  the  load  resistance   P' =  (Q* V*I)/2
.   .   efficiency  of  the  circuit  =   (P' / P  )   =  Q

                                                        which  can  be  made  well  overunity.


I will post it in my forum,
in my opinion normally the energy stored in coils
is limited and the Q will fall to a level under 1
if you draw the power fromm the secondary ?!
Is this true ?

Offline nagarjuna_nallam

  • Newbie
  • *
  • Posts: 2
Re: overunity in resonance ?
« Reply #1 on: August 28, 2005, 05:21:19 PM »
But  stefan,

  when  simulate the  circuit  using software (multisim)  interestingly i   observed  both  voltge  &  current  in  the secondary much  close to  those  values  in  the primary . During simulation  i  observed that the  Q  is  inversly proposional  to resistance in the primary  but  directy  proposional  to the resistance in  the secondary ,i.e. as load resistance  increases  Q of the  circuit increses  (  here  even  though  voltage  increses  to a high value , current  in  both  cicuits  falls  drastically ,  but  still  efficiency  ( = Q)  is  much grater  than  one.).

Offline hartiberlin

  • Administrator
  • Hero Member
  • *****
  • Posts: 8018
    • free energy research
Re: overunity in resonance ?
« Reply #2 on: August 29, 2005, 03:29:47 AM »
Can you please post the output of the simulation as pictures over here ?
Many thanks.
Also the circuit diagramm please.
Pleaseattach it to a reply. Thanks.

Regards, Stefan.

Offline nagarjuna_nallam

  • Newbie
  • *
  • Posts: 2
Re: overunity in resonance ?
« Reply #3 on: September 01, 2005, 07:19:22 PM »
sorry for  the delay stefan. I  am  busy  with my semester work  in  my  college.

                      you  are  asked to send  the  simulation  results  &  circuit  diagram.But  i don't  have  my  own  computer.I am  doing all  this  work at  internet cafe,  this is  also one of the resons  for responding  slowly .

                      stefan  ,i  hope u  understood  my  circuit . once again  i am going to discribe it. I  am using   the  primary  coil  of  a  1:1   transformer as  the  inductor  of  a  series  resonant  circuit  and  across  the  secondary  coil  i  am  putting  my  load .Previously  i  forget  simple  thing  that   even  though  the  current  in  a  series  resonant  circuit  is  in phase  with the  source  voltage ,  it  is  90 degrees  out of  phase  with respect  to  voltage  across  the  inductor/ capacitor.Therefore  same  voltage  &  currents  as  in  the primry  coil  will  appear  in  the secondary ( which  are  90 degrees out of  phase)  and  therfore in the  load. 

                      Now   let us  concentrate  on  the  secondary  circuit.since  secondary  coil  and  load resistance   are  in  series  same  currents  should  flow  through  both  of  them   but  what  about  the  voltages?  since  load  resistance  and  the  secondary   coil  are  the  only  elements  in  the  secondary  circuit  and  more  over  load  resistance  is  across  the  secondary  coil  their  volteges  should  also  same.Here  is  the  contradaction .  voltage  and  current  in  pure  inductor  are  90deg  out  of  phase  but  they  should  be  in  phase in  the  load  resistance.Suppose  if  we  assume  voltages  are  equal  ,how  the  phase  of  current  changes  when it pases  from  inductor  to  resistor  so that  it  is  in  phase  with  the  volage  in the resistor. OR  riversly  If we  assume  current is  same   through  both  inductor  and  resistor  how  voltage across  each  element of   a simple  parellel  combination  will  chage.

              I  hope  you  understand  what  i  am  asking. And  finally  stephan  tommorow  or  day after   tommorow  i will  keep very  intersting  results  & observations   my  simulation,  because  currently i  didn't  brought   them.
                    thank you  for   your  pationce.