# Battery vs Capacitor vs Potential difference

## Homework Statement

I realize I actually don't understand the concept of a battery (potential difference) and a capacitor.

From what I know a battery is something that provides a "drive" for electrons. In other words, take a 90V battery for instance, that just means 90J/C, which means it takes 90J to drive one coulomb of charge across a device. So my question is, where does the word "potential difference" really come in? I know it from textbook memory that it just a, well, a potential difference between two points. But I don't understand what that has to do with a battery (other than units), to me it seems like a Battery is just Electrical Potential, not Potential difference.

Now what about a capacitor? My teacher said to think of a capacitor as a temporary battery, something that stores charges. My question is why do we even have something like that exists? If it is just a temporary battery, why don't we just put a permanent one there? I don't understand, so what if a capacitor hold charges. Why do we even need something that hold charges.

Voltages are always relative to two points. Your 6V battery has 6V delivered to some load only if you hook up both terminals across that load. (Load can be a resistor) If what you're suggesting is true, that a battery needs not be relative to two points, what would you say is the voltage at just one terminal? Remember, if you hook something up to a single terminal in a battery, you get nothing -- and you get nothing, because (more or less) no potential difference exists. A good example of the relativity of voltage comes from birds who sit on 20,000 volt wires without electrocution. Relative to ground, those wires are ~20kV, but across the bird, you have basically 20kv - 20kv = 0V, so they are safe.

Capacitors, that I know of, are primarily used in circuits that have both AC and DC signals. You haven't learned it yet, but a sinusoidal signal sees a sort of 'resistance' (called a reactance or impedance) to a capacitor, and the equation for it is:

$$Z_c = \frac{1}{j 2 \pi f C}$$

where C is the capacitance and f is the frequency of the signal. You can ignore j, for it has implications unimportant to this brief introduction.

So as frequency gets higher (meaning you have, say, a sinusoid alternating very quickly as opposed to one alternating slowly), the impedance that signal sees approaches 0, because you're dividing by frequency. As frequency approaches DC (like a DC battery), impedance approaches infinity (and you see this in the fact that a capacitor charges up and then completely blocks the DC signal from traversing past the capacitor! Note, DC = 0 Hz for frequency, division by zero)

So all this frequency talk may seem irrelevant but it is the lead in for one of the major uses of the capacitor: that it is a filter! If you have a signal v(t) = 50V + sin(2*pi*10000ft) and you only want that quick sinusoid, you can just smash a capacitor on there. It will block the DC and barely block the sinusoid, and on the other end, you'll basically deliver A*sin(2*pi*10000*ft) where A is a minor attenuation factor (like .9 or something). Where would this be useful? Most audio equipment, such as headphones, is designed for only AC signals. The ear hears ~15Hz- 22KHz, so the equipment need not take a 0 Hz (DC) signal and reproduce it for the ear. Doing such may even damage the equipment. Thus, they may use various capacitors and inductors in a complex filter.

What is the other purpose for understanding capacitance? The fact that it is everywhere... even in places we don't want it. Do you ever wonder why we are stuck at around the same frequencies in computers? Part of the problem comes from that 1/f relationship. If we use too quick of a frequency, basically everything becomes shorted by the parallel capacitance, and nothing works.

Maybe an example you would understand would be this: let's say you had a lightbulb modeled by a resistor. Now, all true components have capacitance! Thus, we slap on a capacitor in parallel that is extremely small, maybe 1 x10^-9 Farads. At 60Hz, the impedance is basically infinity... you're dividing by 1x10^-9 = 1x10^9 ohms. What is a resistance in parallel with roughly infinity? It's that resistor. After all, all resistors are in parallel with roughly infinity at all times (air). So you can model its power dissipation and such accurately with JUST the resistor. Ignore that tiny capacitor, for it does nothing. And that is why you will see your calculations for such an experiment work fairly well. Now, let us then say we test that same single resistor with a 1x10^12 Hz sinusoid. We then see we're basically dividing 10^3 in the botton (10^-9 * 10^12 = 10^3) which is 1 milliohm... extremely tiny and much smaller than a conventional resistor. We'll then see, in experiment if you can generate that fast signal, all of the current rushing through that effective capacitance, and the resistive element will see no action. Nothing will work as we planned! In this same way, computer hardware designers are restricted by terminal capacitance in every component they deal with, and if they breach a threshold, everything will basically become a shorted wire, in parallel with 0 ohms.

Hopefully this helps and hopefully it's accurate.

Redbelly98
Staff Emeritus