| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jun 2008 21:04:46 IST
|
|
|
Is the energy stored in a capacitor connected to a battery always half of the work done ( w = QE) by the battery? Pl explain why. Where does the other half go?If wires have negligible resistance, how can the other half of energy go out as heat?
The ckt consists of a capacitor connected to a battery with wires of negligible resistance, nothing else.
|
If the path is beautiful, let us not ask where it leads. But if the DESTINATION is beautiful, let us not ask HOW THE PATH IS. |
|
|
|
24 Jun 2008 21:11:46 IST
|
|
|
energy stored in a capacitor is derived by using calculas n not by da equation,w=qe.da rem half of da energy da is 2 be present actually,is dissipated as heat energy.wenever u charge ur mobile,dont u feel dat it has become hot aft a long time.ceelphone use nothin but capacitors.
|
this reply: 2 points
(with 0 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:13:41 IST
|
|
|
i agree with the answer the other half goes into the thermal energy
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:14:14 IST
|
|
|
We have been discussing this question, anu. Its an ideal case, when wires have negligible resistance.
|
If the path is beautiful, let us not ask where it leads. But if the DESTINATION is beautiful, let us not ask HOW THE PATH IS. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:15:23 IST
|
|
|
heat = I^2 R. But R is zero...
|
If the path is beautiful, let us not ask where it leads. But if the DESTINATION is beautiful, let us not ask HOW THE PATH IS. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:17:44 IST
|
|
|
It is not always that the energy stored in the capacitor is half that provided by the battery. It depends how the capacitor is connected to the battery.
|
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:23:33 IST
|
|
|
Plz explain... when capacitor is connected ulta, eventually battery will provide charge so that the plate connected to positive terminal gets a positive charge and all that...So again the same thing..
|
If the path is beautiful, let us not ask where it leads. But if the DESTINATION is beautiful, let us not ask HOW THE PATH IS. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:34:59 IST
|
|
|
der is no such case as ideal case.here heat is lost not because of da resistance.it is lost because,energ is required 2 store da charge in da capacitor.imagine tryin 2 fill a barrel wit water.1st u hav 2 pour da water in da barrel.dis requires energy ,rt.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 21:57:24 IST
|
|
|
OIC! But isn't that QE!!!
|
If the path is beautiful, let us not ask where it leads. But if the DESTINATION is beautiful, let us not ask HOW THE PATH IS. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 22:25:49 IST
|
|
|
First of all you must know that battery doesn't provide charge.
It accepts charge at lower potential and places it at a higher potential.
Lets take a 10 volt battery connected across a 10uf capacitor. Charge on capacitor will be 100uc. energy provided by battery is 1000uJ energy stored in the capacitor is 500uj. Now we connect the charged capacitor in the opposite way to the battery. Now final potential Energyu is still 500uJ (change in potential energy of capacitor is zero), but energy supplied by battery is 1000uj
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 22:28:21 IST
|
|
|
Energy is lost as heat even in ideal conditions as electrons collide with the negative plate of capacitor and +ve plate of battery and lose their kinetic energy which is radiated as heat.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jun 2008 22:33:08 IST
|
|
|
draw the charge and the potential graph as Q=CV (u can only undst.by drawing graph)
potential on Y-axis.it is a straight line of slope 1/C passing through origin.
consider an element dq
Energy given by battery(of potential V0) is (V0)dq.
But energy required is (corresponding V(v) on Y-axis)dq
V0>v So, there will be energy loss
And clearly u can see the above triangular area(formed by drawing parallel to X-axis) is half of total work done by the battery
|
<SRIRAM.A> on high way of IIT
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
If you are charging with a constant voltage supply then it will be half.
If resistace is zero then in I^2R
I is infinite and R is zero. But integral of I^2Rdt will be finite= half of energy lost by battery.
|
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
2
