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Originally Posted by Meli W.
Maybe it is just b/c my mind is being pulled in 50 million places...
But what exactly happens to a capacitor as it charges and discharges in an RC circuit?
I can't seem to find a website that clearly states it. And it is just something that has been bogging my mind. My physics book doesn't aid either *tear*
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Capacitors are neat... All they are is a set of two parallel plates separated by what is called a dielectric. These dielectrics can be anything that is non-conductive, such as glass or even air. When there is a difference of potential applied to these two plates (such as a battery), the electrons inside the wires and plates will start to move. The electrons will leave the battery and run toward the negative plate, but they stop there because they can't make the "jump" across the plate, and while this is happening the positive end of the battery is sucking electrons out of the positive wire and plate, so a charge builds up on the plates.
When there is no resistor in this circuit, the charge builds up very quickly, because there is nothing to stop the electrons from moving. However, when there IS a resistor in the circuit, it takes time for the electrons to flow through the wires and build up their charge on the two opposite plates. Here comes the math:
Capacitance is defined as being the charge that will build up per unit of voltage on a capacitor, and it is entirely dependent upon the geometry and composition of the capacitor itself. The general formula for capacitance is therefore:
C = q / v, or q = Cv, or v = q / C
where v is the voltage (difference in potential) between the two plates and q is the charge (in Coulombs).
For a capacitor with capacitance C and a voltage source v with a resistor in series R, we can set up the following model:
From Ohm's law, the voltage drop across the resistor is given as:
Vr = I * R,
where I is the current flowing through the resistor, which is also the amount of charge per unit of time that is flowing past it. If you're familiar with calculus, you can write current in the following way:
I = dq/dt
And therefore the voltage drop across the resistor can be written as:
Vr = R * dq/dt
The voltage drop across the capacitor plus the voltage drop across the resistor must be equal to the voltage source, or:
V = Vc + Vr,
where V is the voltage applied by the battery, Vc is the voltage drop across the capacitor, and Vr is the voltage drop across the resistor. Pulling it all together, and remembering that Vc = q / C and Vr = R * dq/dt:
V = q / C + R * dq/dt
You may or may not recognize this as a first order, linear differential equation, which I will try to separate and evaluate using "easy" calculus, though diff eq isn't even a topic most people cover until their second or third year of college...
dq/dt = V / R - q / (C * R)
dq/dt = V / R (1 - q / ((C * R) * (V / R)))
dq/dt = V / R (1 - q / (C * V))
dq / (1 - q / (C * V)) = V / R * dt
Integrating both sides:
-C * V * ln|1 - q / (C * V)| = t * V / R + k <--Constant comes from integration
ln|1 - q / (C * V)| = t * (V / R) / (-C * V) + k / (-C * V)
ln|1 - q / (C * V)| = -t / (R * C) - k / (C * V)
e^(ln|1 - q / (C * V)|) = e^(-t / (R * C) - k / (C * V))
1 - q / (C * V) = e^(-t / (R * C) - k / (C * V))
q = (1 - e^(-t / (R * C) - k / (C * V))) * C * V
q = C * V - e^(-t / (R * C) - k / (C * V))
e ^ (a + b) = e ^ b * e ^ a, so...
q = C * V - e^(-t / (R * C)) * e ^ (- k / (C * V))
Notice that the rightmost expression e ^ (- k / (C * V)) never changes, so it is just a constant. Let's call it 'asdf'. Now,
q = C * V - asdf * e^(-t / (R * C))
q = 0 at t = 0, so:
0 = C * V - asdf * e^(-0 / (R * C))
0 = C * V - asdf * e^0
0 = C * V - asdf => asdf = C * V
So our final equation for the charge on a capacitor in an RC series circuit as a function of time is...
q = C * V - C * V * e^(-t / (R * C))
or q = C * V (1 - e^(-t / (R * C)), which is an equation you might recognize!
PHEW!
Now, let's break down what is actually happening, which is confirmed by the above equation. When the circuit fires up, there is a huge difference between the voltage across the capacitor (0v) and the battery's voltage. This makes lots of current flow. However, as the current flows, the capacitor gets more and more charge on it, which increases the voltage across the capacitor (V = q / C), which DECREASES the difference between the capacitor's voltage and the battery voltage, which DECREASES the current. Thus, as t->infinity, the current approaches zero as the charge approaches some final value!
When the cap discharges, there is no battery involved. The current will be proportional to the voltage drop that is still across the capacitor, and it will decrease as the charge becomes less and less. It's like filling up a bucket and poking a hole in the bottom. When the water gets lower, the rate at which it comes out is lowered, too. If you still have any questions, just reply and ask. For now, I need sleep...