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Charging and discharging capacitors - current time graph

by jsmith613
Tags: capacitors, charging, current, discharging, graph, time
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jsmith613
#1
Apr3-12, 08:23 AM
P: 614
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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gneill
#2
Apr3-12, 09:32 AM
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Think in terms of the potential difference that is driving the current. How much of the potential appears across the resistor over time?
jsmith613
#3
Apr3-12, 11:42 AM
P: 614
Quote Quote by gneill View Post
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

gneill
#4
Apr3-12, 11:46 AM
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P: 11,783
Charging and discharging capacitors - current time graph

Quote Quote by jsmith613 View Post
well p.d increases over time
In both cases? For the resistor?
jsmith613
#5
Apr3-12, 12:52 PM
P: 614
I think so...
gneill
#6
Apr3-12, 01:05 PM
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P: 11,783
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?
jsmith613
#7
Apr3-12, 01:09 PM
P: 614
Quote Quote by gneill View Post
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
gneill
#8
Apr3-12, 01:46 PM
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P: 11,783
Quote Quote by jsmith613 View Post
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.
jsmith613
#9
Apr3-12, 03:11 PM
P: 614
Quote Quote by gneill View Post
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 :(
gneill
#10
Apr3-12, 03:19 PM
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Quote Quote by jsmith613 View Post
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?
jsmith613
#11
Apr3-12, 04:18 PM
P: 614
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)
gneill
#12
Apr3-12, 04:39 PM
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Quote Quote by jsmith613 View Post
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...
jsmith613
#13
Apr3-12, 04:43 PM
P: 614
Quote Quote by gneill View Post
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?
gneill
#14
Apr3-12, 04:49 PM
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Quote Quote by jsmith613 View Post
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.
gneill
#15
Apr3-12, 04:56 PM
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Attached Thumbnails
Fig1.gif  
jsmith613
#16
Apr3-12, 05:30 PM
P: 614
Quote Quote by gneill View Post
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
gneill
#17
Apr3-12, 05:40 PM
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Quote Quote by jsmith613 View Post
well I am used to using a two way switch :S
Does that change the essential functionality of the two scenarios?
jsmith613
#18
Apr3-12, 05:53 PM
P: 614
Quote Quote by gneill View Post
Does that change the essential functionality of the two scenarios?
well no but it does change the circuit diagram


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