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Rumieus
Joined: 07 Jul 2008
Posts: 20
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 19 Aug 2008 18:25 capacitor discharge path |
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when the switch move to position C, how the voltage in capacitor flow? from A to C? or from B to ground? can someone explain. Tq very much.
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Prototyp_V1.0
Joined: 03 Apr 2007
Posts: 62
Helped: 9
Location: Norway
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| 19 Aug 2008 18:46 Re: capacitor discharge path |
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I didn't know that voltage could "flow". I guess you mean that current flows. It flows the whole way round through R2, doesn't affect the battery, zero amps there.
Please save the pic as png next time. Jpg pictures is not suitable for chematics. It got a redicilus filesize and looks ugly. A png version would probably have a filesize below 10 kb.
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Stewie911
Joined: 12 Aug 2008
Posts: 12
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| 20 Aug 2008 1:58 Re: capacitor discharge path |
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The cap will act like a voltage source with its positive terminal at A.
The current actualy flows from C to A, but for all practical calculations we say it flows from A to C. The same is true for the battery, the current actually flows from + to -, but we say it flow from - to +(called Conventional current).
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insrusingh
Joined: 08 Jul 2008
Posts: 16
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| 20 Aug 2008 5:17 capacitor discharge path |
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| If you consider like curent is in opposite the direction of flow of electrons.So, when switch moves to point C then across capacitor A us at positive potential and B is at ngative potential .so in closed loop current flows from A to C but if you RE TALKING ABOUT convenional current it flows from B to A.
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subharpe
Joined: 09 Jan 2008
Posts: 163
Helped: 21
Location: Bangalore,India
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| 21 Aug 2008 9:10 Re: capacitor discharge path |
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| Initially when the battery was connected to the capacitor through R1 for a long time the capacitor will be charged to the voltage of the battery 12V. Now when theswitch position is changed so as to allow the capacitor to discharge through R2 the initial voltage at the point C wil be 12V. Then the voltage across the capacitor will decrease esponentially with the relation V(t) = 12exp(-t/[R2*C]). The current through the loop will decrease from 12V/R2 with the expresion V(t)/R2.
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saruman1983
Joined: 12 Oct 2005
Posts: 25
Helped: 4
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| 21 Aug 2008 12:19 Re: capacitor discharge path |
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1)There is no ground in the schematic
2)We say that the electric current flows the same direction as the positive charge carriers do and the opposite of that which negative charge carriers do. In the given schematic that is from A to C to B (when the switch moves to C the source V1 and resistor R1 do not affect the situation). Electrons are negative charge carriers, so they will flow to the opposite of the A to C to B direction (that is B to C to A), while holes which are positive charge carriers travel the same way as the current. In general one can consider the electric current flowing to the opposite direction of that which electrons do
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