Home Page News Search Contact Us Language Bar فارسی English
 
 
 
Electrical riddle no.21 - Missed Energy
We connect both of them via an electrical switch as following diagrams. According basic electrical equation half of stored energy misses after that switching.

Where dose missed energy go? What is your opinion?

Wa=1/2CV2


Wb= 1/8 CV2+1/8CV2=1/4CV2



Author : Hamid - From: IRAN
 
#1
Tue, September 9, 2008 - 09:04
After equalization , new voltage won't be V/2 , it is 0.707   V    
Author : AMIR KHASHAYAR - From: IRAN
 
#2
Tue, September 9, 2008 - 11:36
after closing the switch and equalization the voltage won't be V/2 . It is 0.707 V . Why it should be V/2 ? 
Author : AMIR KHASHAYAR - From: IRAN
 
#3
Tue, September 9, 2008 - 17:33
Please refer to simple equation of capacitors voltage(v=q/c), and survival of charges law. 
Author : Hamid - From: IRAN
 
#4
Wed, September 10, 2008 - 07:17
At first electric field inside new capacitor is zero . after closing of switch the field of new capacitor is gradually increased. After equalization field in both capacitors are equal and it's E/2. for moving of electrical element inside field we need energy . The lost energy is cosumed for transferring of electrical element from negative pole to positive pole . 
Author : GLAVIZH - From: IRAN
 
#5
Sun, September 14, 2008 - 06:44
Lost energy has been converted to electromagnetic radiation . transferring of charge from charged capacitor to new one isn't static process .It's dynamic and coincident with oscillation . this process generates radiation and gives heat. 
Author : RONAK - From: IRAN
 
#6
Thu, November 13, 2008 - 14:33
Faraday told us that an electric current generates a magnetic field and a changing magnetic field can create a current in a conductor.  Maxwell refined this and we know that a changing electrostatic field generates a magnetic field and a changing magnetic field generates an electrostatic field.

If we have 2 identical ideal capacitors connected in an ideal circuit as described without any resistance and close the switch.......

The first capacitor starts to discharge and the electrostatic field in the capacitor decreases in strength creating a magnetic field.  As the capacitor continues to discharge, the rate of decrease of the electrostatic field begins to decrease as well and the magnetic field begins to collapse.  The collapsing magnetic field creates it's own electrostatic field which reinforces the one in the capacitor that is decreasing.

A similar but opposite magnetic field is building and collapsing in the second capacitor with the result that by the time the magnetic fields are gone, all the charge has moved to the second capacitor.  The cycle then runs in reverse and we have oscillation.

At the point of the cycle where there are equal amounts of charge in each capacitor, there is energy stored in the magnetic fields equal to the energy stored in the electrostatic fields.

In the real world, energy is lost through radio wave emission due to high frequency oscillations and heat losses due to resistance.  These losses cause the oscillations to decay to nothing. 
Author : Stephen Byrne - From: Australia
 
#7
Thu, November 13, 2008 - 23:42
1- First part of your answer is interesting, i must think about it.
2-Regarding to real part of you discussion, i can say,the high frequency oscilation is not possible in mentioned circuit but ohmic losses in arc path of switch can be occure. 
Author : Hamid - From: Iran
 
Submit Your Answer

 
 
Change Language :