Capacitor Discharge: Deriving the Correct Expression for Potential Difference

In summary, the conversation discusses deriving an expression for the potential difference over a capacitor at any time t, using Kirchhoff's voltage law and considering the direction of current flow. The correct equation is V_{c} (t) = V_{initial} e^{\frac{-t}{RC}} where V_{initial} = \frac{Q}{C}. The mistake in the incorrect equation is not taking into account the direction of current flow.
  • #1
Ahmedbasil
8
0

Homework Statement


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. Homework Equations and theorems
[tex]I_{c} = C\frac{dV_{c}}{dt}[/tex]
[tex]V = IR[/tex]

*Kirchhoff's voltage law

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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  • #2
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.
 
  • #3
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.
 
Last edited:
  • #4
Ahmedbasil said:
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.
 
  • #5
Owh, okay, I assumed it is the current OUT of the capacitor. I see :). My bad, and thank you very much.
 

1. What is a capacitor?

A capacitor is an electronic component that stores and releases electrical energy. It is made up of two conductive plates separated by an insulating material, called a dielectric. When a voltage is applied, one plate accumulates a positive charge while the other accumulates a negative charge, creating an electric field between them.

2. How does a capacitor discharge?

A capacitor discharges when the stored electrical energy is released. This can happen in a controlled manner through a circuit, or spontaneously due to the natural leakage of the dielectric material. When a discharge occurs, the electric field between the plates collapses and the charges flow from one plate to the other, resulting in a flow of current.

3. What factors affect the discharge of a capacitor?

The rate of discharge of a capacitor is influenced by several factors, including the capacitance (size) of the capacitor, the voltage across the capacitor, and the resistance of the circuit. A larger capacitor will take longer to discharge, while a higher voltage and lower resistance will result in a faster discharge.

4. Can a capacitor be discharged completely?

No, a capacitor cannot be discharged completely. Even after a discharge, there will still be a small amount of residual charge remaining on the plates. This is due to the natural leakage of the dielectric material, which cannot be completely eliminated.

5. What are some practical applications of capacitor discharge?

Capacitor discharge has several practical applications, including in camera flashes, strobe lights, and defibrillators. It is also used in power supplies to smooth out fluctuations in voltage and in ignition systems for internal combustion engines. Additionally, capacitor discharge is essential for the proper functioning of many electronic devices and circuits.

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