Thanks for this explanation ! I assume if we plot a Voltage vs Electrons transferred graph, we can see the first point (first electron) has 0V, and the last point (last electron) has 9V, it will look like a linear ascending line from 0 to 9V. This forms a triangle which is half of total charge (Q) * voltage (9V)
Yes, and this example suffers by showing a 9V battery, which implies that as soon as the switch closes the capacitor plates are at 9V. It would be better to replace the battery with a current source, and show it charging the cap from zero volts to 9V.
Where does half the energy go? If you were to go by the definition of potential difference (work done to move 1 C charge across) then we find that V=W/Q and we know that W = Q V/2. But if we were to sub in a value for V ie. W = Q W/ Q 2 (Q gets cancelled) W = W/2 or 1= 1/2 and that's not true so how is the energy magically halved? Where does half the energy go? Heat is obviously not the answer but where else could this 1/2 go?
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I can't thank you enough .All your lessons are extremely helpful!
Sir very nice explaination . Thanks a lot ... LOTS OF LOVE FROM INDIA
Thanks for the comment and and you are most welcome from Berlin Germany.
sir, your videos had saved me long hours of reference reading. thanks a lot 👍👍
You are most welcome, best wishes.
Excellently explained 👍👍
Thank you, glad you think so!
this was very useful! thanks!
You're welcome, glad it was helpful!
sir ur explanation s trly very sweet and simple it has helped me lot to learn difficu concepts in easy way
thank u so much
Thanks. Yes I try to get right to it. No reason to waste time.
Very articulate and clear!
Extremely Helpful sir Was struggling to understand this ;-; but it makes sense :)
Great, hope you get it.
Thanks for this explanation ! I assume if we plot a Voltage vs Electrons transferred graph, we can see the first point (first electron) has 0V, and the last point (last electron) has 9V, it will look like a linear ascending line from 0 to 9V. This forms a triangle which is half of total charge (Q) * voltage (9V)
Yes, and this example suffers by showing a 9V battery, which implies that as soon as the switch closes the capacitor plates are at 9V. It would be better to replace the battery with a current source, and show it charging the cap from zero volts to 9V.
Awesomeeee.thanks so much.
You bet!
Very nice sir
Thanks a lot
Great video. Now my text makes more sense!
Awesome!
thank you so much
Excellent! Thanks
would like to ask so how is the total energy taken from the battery during the charging process is different to the total energy of the battery?
Very inspiring thanks bro
Tungod jud ning maam Ruiz maong naa ko diri
Nice to see you on my channel. Thanks for watching!
Hi sir , Is there any capacitor that can stores energy for the day or days?
if they are in an electric circuit, yes....otherwise they lose their charge
Where did the "1/2" come from?
That is a great question, did I not explain it in the video. I know I did somewhere.
I have a 12 volt fan can i put capacitor and what capacitor i put
which combination store more energy????
Educative
Thanks!
Where does half the energy go?
If you were to go by the definition of potential difference (work done to move 1 C charge across) then we find that V=W/Q and we know that W = Q V/2.
But if we were to sub in a value for V ie.
W = Q W/ Q 2 (Q gets cancelled)
W = W/2 or
1= 1/2 and that's not true so how is the energy magically halved? Where does half the energy go? Heat is obviously not the answer but where else could this 1/2 go?
Am the 600 like guy
❤