# Voltage across Capacitor equation derivation. Pls help, much appreciated

Watch
Announcements
#1
so i came across this on I want to study engineering. I have drawn arrows at the point where they times both sides by -1 so that a decay exponential ie e^-1/RC is formed. My question is, mathematically what makes you do this. It could turn into an exponential increase if you don't right? so is it done out of intuition or math?? THANKS
0
5 years ago
#2
You will get the exact same formula even if you do not multiply by -1. I think it might be because it makes the LHS integral a tiny bit nicer (though it really makes no difference in my opinion):

Now is probably the time where a physicist comes in an tells me it is not as simple as the maths suggests haha.
0
5 years ago
#3
(Original post by 16Characters....)
You will get the exact same formula even if you do not multiply by -1. I think it might be because it makes the LHS integral a tiny bit nicer (though it really makes no difference in my opinion):

Now is probably the time where a physicist comes in an tells me it is not as simple as the maths suggests haha.
Nah, you got it in one.

The equation tells us the charge stored in an initially isolated capacitor (Vc = Vo) will exponentially decay at a rate governed by the product between the capacitance and current path resistance.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (226)
55.67%
I don't have everything I need (180)
44.33%