Capacitors in Circuits

Recall that the capacitance of a parallel plate capacitor can be given in terms either of the construction details of the capacitor (C = ke0A/d) or in terms of the charge and voltage applied to the capacitor, or C = Q/V.  Now, consider the circuit below where we place two capacitors, C1 and C2 in series with a battery.

The left plate of C1 will be at the same potential as the + side of the battery and will build + charges, and the right side plate of C2 will be at the same potential as the - side of the battery with - charges. The right side of C1 will then build up a negative charge and by induction the left side of C2 will build up a positive charge.

V = V1 + V2 and Q is the same on all capacitors. From above,

This is the equation for finding Equivalent Capacitance in a circuit for series capacitors. Note that it looks like the formula for finding equivalent resistance for parallel resistors.

Now, consider a case of 2 capacitors in parallel as shown:

Here, the voltage drop across each capacitor is equal to V. Additionally, since the current will branch off between the two capacitors, the charge will be distributed between C1 and C2, such that Qtot = Q1 + Q2. Thus, we can say that CV1 + CV2 = CV, or

C = C1 + C2

Again, this is opposite the pattern that we find for resistors in parallel.

When capacitors and resistors are networked together in an RC Circuit as shown to the right, when the switch is shown, current will immediately start to flow through the circuit. As charge accumulates on the capacitor, the potential difference across it will eventually equal the potential difference across the battery, and at this point, no charge will flow. The potential difference across the capacitor increases with time, and this is an exponential function given by the equation

V= Vo (1 - e-t/RC), where RC is the time constant of the circuit. A graph would look like this:

The time constant is a measure of how quickly the capacitor is charged. If the battery is removed, when the switch is closed now, the voltage will decay away as current passes through the resistor. The exponential
decay curve is given by

V = Voe-t/RC

and looks like:

Here is a quick problem:

Four 2mF (microfarad) capacitors are connected in series, what is the equivalent capacitance? What would the equivalent capacitance be if they were in parallel?

Answer:
In series  1/Ceq=1/2+1/2+1/2+1/2
1/Ceq=2
1/2=Ceq
Ceq=.5mF

In , parallel Ceq=2+2+2+2
Ceq=8mF

For some additional reading: http://www.tpub.com/neets/book2/3e.htm

The NTNU (National Taiwan Normal University) Virtual Physics Laboratory provides excellent resources and I'm a big fan. Click here for an applet that shows the transient nature of capacitors and charge in DC circuits.

For Practice Problems, Try: Giancoli Multiple Choice Practice Questions (Go ahead - try a few.)