View Full Version : RC Circuits

Melissa Nute
01-24-2004, 09:33 PM
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*

Caleb Fulton
01-25-2004, 01:42 AM
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*

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!


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...

01-25-2004, 03:15 AM
Ah, great question! It's great to hear some more complex aspects of circuits and circuit analysis. Alright, so we can basically break things up in to two main categories...resistive circuits and RC/RL circuits. If we take a nodal/loop equation for a resistive circuits, we get algebra equations (often we use matrices to solve these). Now for RC/RL circuits we have interconnections of sources, resistors, capacitors, and inductors. Inductors and capacitors have differential or intgral voltage and/or current relationships. Since your question pertains specifically to RC capacitors, I'll just talk about that. So, when we have a Capacitor, we have an integral relationship. This is because it takes time to charge up and discharge the capacitor, which brings in calculus (YAY!!!). When one capcitor is there in a circuit with a resisor and sources we get a first order RC circuit. Here Lamda will be represented by ' L '. These circuits will have an exponential response which means it will be proportional to A + Be^Lt for constants A, B, and L. So, if we have a source and a resistor in series with a capacitor we will hvae some initial capcitor voltage Vc(0-) where the 0- describes the instant just before zero. If we take a loop equation here, we get Vs(t) = RIc(t) + Vc(t). Because Ic(t) = C*dVc(t)/dt and if we substitute this into the first eqn, we get Vs(t) = RC * dVc(t)/dt + Vc(t). or... dVc(t)/dt + 1/RC*Vc(t) = 1/RC*Vs(t). So, this equation shows us that the derivative of the capacitor voltage plus 1/RC times the capacitor voltage equals 1/RC times the source voltage. So, in the circuit that has a voltage source and a resistor in series with a capacitor, we can show the capacitor current Ic(t) as the differential equation:
dIc(t)/dt + 1/RC*Ic(t) = 1/R * dVs(t)/dt.

So I have thrown a lot of equations and stuff, kinda complex stuff. Essentially it is based on differential equations which is some upper level math. Basically, a capacitor is going to be used in order to create some type of signal. I guess if you take a look at a tv, this can give a good general example. In the TV transmitter u have a signal which is essential in creating the image on the screen. If you have an oscilloscope a timing signal is used that acts as a tibe base which lets you see measured input signals as a function of time. When you hook this up to the TV, you will get a certain output on the oscilloscope. In the perfect conditions, the voltage will increase in a linear fashion with the time until it gets to a region where it will all of a sudden go to zero, which starts this chain process over again. This voltage region corresponds to a fixed unit of time. The linear voltage goes up and then acts as an electronic "second" hand, ticking off the smaller units of time. The linear increase in the voltage is approximated by the linear part of an exponential response in the RC circuit. When the voltage measured across the capacitor reaches a certain range, the capacitor is quicky discharged to zero. Once this happens, another switch reactivates the circuit and starts the process over. Basically, the capacitor will set a certain type of output which we want to use to reproduce. Via differential equations, we can calculate certain aspects of the RC circuit.

I know that this was rather long, lengthy and maybe not totally clear. I hope I was able to help even a little bit and at least gave you a little clearer answer/background. Yes it's complex, but once you really take a look at it, it's not tooo bad. Let me know if you have any more questions.

Jay Lundy
01-25-2004, 04:02 AM
Sorry, I didn't read the above 2 posts, so I might repeat something, but I'll try to keep this short and intuitive.

As electrons travel from one plate of the cap to the other (through the battery, not the dialectric), a voltage builds up across the plates. That voltage acts against the voltage created by the battery. So, when the cap has no charge, voltage is greatest and therefore current is greatest (V=IR). When there is a higher current, the voltage across the cap rises faster (more electrons travel from one plate to another). Then as the voltage across the cap rises, current decreases and therefore the rate at which it charges decreases. Eventually the voltage across the cap is so close to the battery voltage that it is basically 0 and there is no (well, very little) current. Hence you get exponential curves.

If you do the math, you get T= RC where T is the timing constant, R is resistance in ohms (more ohms = less current = more time for constant C), and C is capacitance in farads.

This applet might help too: http://www.phy.ntnu.edu.tw/java/rc/rc.html