05 Nov '09 15:31>
My electricity teacher asked me this question:
Imagine a circuit with two equal capacitors, two switches and a DC source
(The o's don't mean anything, the forum just won't let me put blanks in between the 'wires'😉
|----| |------|
|ooooooooo|
|ooooooooo /
|----| |------|
|ooooooooo|
|ooooooooo|
|-----DC----|
Now, if the source is turned on the bottom capacitor will charge up to Q=CV
This means an energy of 1/2CV² is stored in that capacitor
The upper capacitor won't do anything.
Now if the DC is source is removed by means of a switch and if then the other switch is closed , the bottom capacitor will partly discharge into the upper capacitor.
|----| |------|
|ooooooooo|
|ooooooooo|
|----| |------|
|ooooooooo|
|ooooooooo /
|-----DC----|
There will now be Q/2 charge on each capacitor and a voltage of V/2 over each one. That means a total energy of 2*1/2*C*(V/2)²= 1/4C*V
The energy has been halved!
Where did the other half go?
Imagine a circuit with two equal capacitors, two switches and a DC source
(The o's don't mean anything, the forum just won't let me put blanks in between the 'wires'😉
|----| |------|
|ooooooooo|
|ooooooooo /
|----| |------|
|ooooooooo|
|ooooooooo|
|-----DC----|
Now, if the source is turned on the bottom capacitor will charge up to Q=CV
This means an energy of 1/2CV² is stored in that capacitor
The upper capacitor won't do anything.
Now if the DC is source is removed by means of a switch and if then the other switch is closed , the bottom capacitor will partly discharge into the upper capacitor.
|----| |------|
|ooooooooo|
|ooooooooo|
|----| |------|
|ooooooooo|
|ooooooooo /
|-----DC----|
There will now be Q/2 charge on each capacitor and a voltage of V/2 over each one. That means a total energy of 2*1/2*C*(V/2)²= 1/4C*V
The energy has been halved!
Where did the other half go?