Discharge of a capacitor


by Ahmedbasil
Tags: capacitor, discharge
Ahmedbasil
Ahmedbasil is offline
#1
Nov21-11, 04:40 AM
P: 8
1. The problem statement, all variables and given/known data
You have a capacitor, of capacitance C farads, with charge Q coulombs. It is connected in series with a resistor of resistance R ohms. Derive an expression for the potential difference over the capacitor at any time t.


2. Relevant equations and theorems
[tex]I_{c} = C\frac{dV_{c}}{dt}[/tex]
[tex]V = IR[/tex]

*Kirchhoff's voltage law


3. The attempt at a solution
Using KVL:

[tex]V_{c} - I_{c} R = 0[/tex]
[tex]V_{c} = RC\frac{dV_{c}}{dt}[/tex]
[tex]\frac{1}{RC} dt = \frac{1}{V_{c}} dV_{c}[/tex]

then:

[tex] V_{c} (t) = V_{initial} e^{\frac{t}{RC}}[/tex]

where

[tex]V_{initial} = \frac{Q}{C}[/tex]

3. My concern

as t approaches infinity the potential difference over the capacitor also approaches infinity. This is definitely not right - the capacitor is discharging. Every textbook/website I look at comes up with the equation:

[tex] V_{c} (t) = V_{initial} e^{\frac{-t}{RC}}[/tex]

For the life of me I cannot figure out what I'm doing wrong. I know what's supposedly wrong with my solution, but I cannot see the mathematical proof of the minus sign on the power of e.

I'd greatly appreciate any help or insight anyone could give.
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gneill
gneill is offline
#2
Nov21-11, 06:28 AM
Mentor
P: 11,416
It's a current direction issue. dV/dt will be negative for the capacitor, right? (It's discharging, so its voltage is dropping). But the current in your circuit you've defined to be positive flowing out of the capacitor. So you should start with Ic = -C dVc/dt.
Ahmedbasil
Ahmedbasil is offline
#3
Nov21-11, 07:19 AM
P: 8
I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t another variable negative. The equation should be:

[tex]I_{c} = C |\frac{dV_{c}}{dt}|[/tex]

Anyway, thanks very much for your help.

gneill
gneill is offline
#4
Nov21-11, 07:26 AM
Mentor
P: 11,416

Discharge of a capacitor


Quote Quote by Ahmedbasil View Post
I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t another variable negative. The equation should be:

[tex]|I| = C |\frac{dV_{c}}{dt}|[/tex]

Anyway, thanks very much for your help.
The defining equation for the capacitor does take direction into account. In the definition, the current direction is defined to be INTO the capacitor, and resulting voltage change is positive. Thus, when the voltage change is NEGATIVE, the current will come OUT of the capacitor.
Ahmedbasil
Ahmedbasil is offline
#5
Nov21-11, 07:27 AM
P: 8
Owh, okay, I assumed it is the current OUT of the capacitor. I see :). My bad, and thank you very much.


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