## Charging and discharging capacitors - current time graph

1. The problem statement, all variables and given/known data
why is the current-time graph for a charging AND discharging capacitor the same?

2. Relevant equations

3. The attempt at a solution
Q=It
so for a discharging capacitor as time goes on the charge stored decreases so current decreases

BUT for a charging capacitor charge increases so current should increase??
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 Recognitions: Homework Help Think in terms of the potential difference that is driving the current. How much of the potential appears across the resistor over time?

 Quote by gneill Think in terms of the potential difference that is driving the current. How much of the potential appears across the resistor over time?
well p.d increases over time

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## Charging and discharging capacitors - current time graph

 Quote by jsmith613 well p.d increases over time
In both cases? For the resistor?
 I think so...
 Recognitions: Homework Help Sketch the PD across the capacitor for the charging case. You should be able to infer the PD across the resistor using KVL for the circuit (battery, resistor, capacitor). What do you find?

 Quote by gneill Sketch the PD across the capacitor for the charging case. You should be able to infer the PD across the resistor using KVL for the circuit (battery, resistor, capacitor). What do you find?
well in one case it is increase (charging) and the second case it is decreasing (discharging) expotentially

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 Quote by jsmith613 well in one case it is increase (charging) and the second case it is decreasing (discharging) expotentially
To what are you referring by 'it'? Let us be clear; there are three potential differences to be concerned with in the circuit. One is the PD of the battery which is fixed (a constant), the second is the PD across the resistor, and the third is the PD across the capacitor.

 Quote by gneill To what are you referring by 'it'? Let us be clear; there are three potential differences to be concerned with in the circuit. One is the PD of the battery which is fixed (a constant), the second is the PD across the resistor, and the third is the PD across the capacitor.
in charging case, pd across capacitor increases and in discharging case pd across capacitor decreases

no idea about resistor, sorry :(

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 Quote by jsmith613 in charging case, pd across capacitor increases and in discharging case pd across capacitor decreases no idea about resistor, sorry :(
How do the potential differences around the loop add up?
 V = V1 + V2 so I would presume the emf across the resistor falls as p.d across capacitor increases (and then increases as p.d across capacitor decreases)

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 Quote by jsmith613 V = V1 + V2 so I would presume the emf across the resistor falls as p.d across capacitor increases (and then increases as p.d across capacitor decreases)
That's nearly right. Certainly the p.d. across the resistor falls as the p.d. across the capacitor rises when it's charging. But what about the case where the capacitor is discharging? In that case the battery is removed and replaced with a piece of wire (the circuit is different). The capacitor starts out with some initial p.d. across it, and the resistor is connected directly across the capacitor. So the resistor starts with the same p.d. as the capacitor...

 Quote by gneill That's nearly right. Certainly the p.d. across the resistor falls as the p.d. across the capacitor rises when it's charging. But what about the case where the capacitor is discharging? In that case the battery is removed and replaced with a piece of wire (the circuit is different). The capacitor starts out with some initial p.d. across it, and the resistor is connected directly across the capacitor. So the resistor starts with the same p.d. as the capacitor...
hold on...when its charging the resistor is not involved
it is only involved during discharging

surely your approach is wrong?

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 Quote by jsmith613 hold on...when its charging the resistor is not involved it is only involved during discharging surely your approach is wrong?
The resistor must be there in both cases. Otherwise you've got an unrealistic circuit.
 Recognitions: Homework Help Attached Thumbnails

 Quote by gneill The resistor must be there in both cases. Otherwise you've got an unrealistic circuit.
well I am used to using a two way switch :S

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 Quote by jsmith613 well I am used to using a two way switch :S
Does that change the essential functionality of the two scenarios?

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