Capacitors -- charging and discharging

  • Thread starter Hannah7h
  • Start date
  • #1
Hannah7h
40
0
So, is this true for when a capacitor charges through a fixed resistor (as shown in the image below, when the switch is closed to 1)- the potential difference across the resistor exponentially decreases to zero and the potential difference across the capacitor exponentially increases from zero to equal the voltage across the battery (power supply), when the capacitor is fully charged?
Screen Shot 2016-06-06 at 18.54.30.png
 

Answers and Replies

  • #2
cnh1995
Homework Helper
Gold Member
3,448
1,148
So, is this true for when a capacitor charges through a fixed resistor (as shown in the image below, when the switch is closed to 1)- the potential difference across the resistor exponentially decreases to zero and the potential difference across the capacitor exponentially increases from zero to equal the voltage across the battery (power supply), when the capacitor is fully charged?
View attachment 101750
Yes.
 
  • #3
Hannah7h
40
0
Yes.

Ok, so now if the switch is moved to position 2 and the capacitor is now discharging. The potential difference of the capacitor will now decrease exponentially, and so does the potential difference across the fixed resistor- why does the p.d across the fixed resistor also decrease?
 
  • #4
David Lewis
842
251
Also, when switch is in position 1, you could say voltage across the cap rises asymptotically to battery voltage.

When switch is in position 2, you can then think of the capacitor as a power supply -- where the voltage drops as you draw current out of it.
 
  • #5
cnh1995
Homework Helper
Gold Member
3,448
1,148
Ok, so now if the switch is moved to position 2 and the capacitor is now discharging. The potential difference of the capacitor will now decrease exponentially, and so does the potential difference across the fixed resistor- why does the p.d across the fixed resistor also decrease?
Capacitor acts as a source while discharging and voltage across capacitor is equal to the voltage across resistor. Since capacitor voltage is decreasing, voltage across the fixed resistor is also decreasing.
 
  • #6
cnh1995
Homework Helper
Gold Member
3,448
1,148
In other words, resistor does not store any charge. Hence, voltage across the resistor decreases while discharging of the capacitor.
 
  • #7
Hannah7h
40
0
Capacitor acts as a source while discharging and voltage across capacitor is equal to the voltage across resistor. Since capacitor voltage is decreasing, voltage across the fixed resistor is also decreasing.

Ohhh I see that makes sense, just one more question why in: VC= -VR is the p.d across the resistor quoted as a negative value?
 
  • #8
Hannah7h
40
0
Also, when switch is in position 1, you could say voltage across the cap rises asymptotically to battery voltage.

When switch is in position 2, you can then think of the capacitor as a power supply -- where the voltage drops as you draw current out of it.

Ah ok yes this makes sense now
 
  • #9
cnh1995
Homework Helper
Gold Member
3,448
1,148
Ohhh I see that makes sense, just one more question why in: VC= -VR is the p.d across the resistor quoted as a negative value?
It follows from KVL i.e. Vc+Vr=0. Going along the direction of current, you'd see a drop in potential across the resistor and a gain of potential across the capacitor.
 
  • #10
Hannah7h
40
0
It follows from KVL i.e. Vc+Vr=0. Going along the direction of current, you'd see a drop in potential across the resistor and a gain of potential across the capacitor.

Ok yeah makes sense, so when the capacitor is discharging the p.d across the capacitor increases so the p.d across the resistor decreases in order to = zero. And then when the capacitor discharges VC= -VR as both the p.d across the capacitor and p.d across the resistor decrease to zero.
 
  • #11
cnh1995
Homework Helper
Gold Member
3,448
1,148
Ok yeah makes sense, so when the capacitor is discharging **charging** the p.d across the capacitor increases so the p.d across the resistor decreases in order to = zero
And then when the capacitor discharges VC= -VR as both the p.d across the capacitor and p.d across the resistor decrease to zero.
Right.
 
  • #12
Hannah7h
40
0
Right.

Sorry my mistake! and ok good, thanks for the help!
 
  • #13
cnh1995
Homework Helper
Gold Member
3,448
1,148
Sorry my mistake! and ok good, thanks for the help!
You're welcome!
 
  • #14
lychette
413
72
Also, when switch is in position 1, you could say voltage across the cap rises asymptotically to battery voltage.

When switch is in position 2, you can then think of the capacitor as a power supply -- where the voltage drops as you draw current out of it.

Rises exponentially is the correct and precise term!
'Asymptotically' is unnecessary and misleading.
Some 'asymptotic' curves are not exponential.
If the charging/discharging curves of capacitors were described as 'asymptotic' in an exam answer it would be marked wrong !!
Take care with terminology!
 

Suggested for: Capacitors -- charging and discharging

Replies
7
Views
2K
Replies
21
Views
10K
Replies
8
Views
3K
Replies
50
Views
9K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
789
  • Last Post
Replies
3
Views
2K
Replies
19
Views
944
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
820
Top