Capacitors
A capacitor stores charge, equal amounts of positive and negative charge. (Equal amounts is not demanded by the theory, but in practice, that's the way it is.) Usually the charge is stored for a fraction of a second-- for example to smooth out voltage fluctuations or as a component in an oscillator circuit. Longer time storage examples include a capacitor that is used to power the flashbulb of a camera and a capacitor used to keep an electronic circuit running while the battery is replaced. (If the directions say something like "you have two minutes to replace the battery or else all memory will be lost," that is a good clue that it is a capacitor that is running the thing while the battery is out.) People are working on a big leap forward in capacitor technology so that electric cars could be powered by capacitors rather than batteries, but maybe this won't work out. It seems far-fetched.
Normally a capacitor is two conductors separated by an insulator. If you put
+q charge on one and -q on the other, the voltage difference will be
proportional to q. We write
q = CV, and this defines the capacitance C. If you put +q on one plate, the other plate will suck in
-q from somewhere. If it can't, the charges on both plates will arrange
themselves in a minimum energy configuration which depends on the geometry, and
this is a difficult problem. Think +q and -q in all practical
cases.
The unit of capacitance is the farad, which is a Coulomb/volt. In ancient
times I would tell students that a one farad capacitor would need to be the
size of the classroom. No more- you can buy one (called a supercapacitor)
about the size of a few quarters for five bucks or so. More commonly, the unit
is microfarad
(m f = 10-6 f) or picofarad (pf = 10-12). Never use millifarad, or the capacitance police will come with a
charged capacitor and zap you.
You are a capacitor. Get an electrical tester that consists of a tiny neon bulb with a resistor in series with it. Hold one end and connect the other end to the "hot" wire in an electrical outlet. The bulb lights up, due to the fact that you charge up alternately plus and minus, thus there is a small alternating current in the bulb. It is such a tiny current that you will not feel it. When you are charged, there is an induced opposite charge nearby in anything grounded, so in a way you are teaming up with the surroundings to form a capacitor.
In the first paragraph I mentioned that a capacitor can smooth out voltage fluctuations. The picture below illustrates how this works. The picture on the left is a water pump which pumps in spurts clockwise around the loop. The thin part of the tube provides resistance to flow. In the center there is a rubber membrane which stretches downward when the pressure above is high, then when the pressure goes down it comes back up, pushing water up, helping to maintain the flow through the "resistor." Similarly the picture on the right shows a generator which generates current in spurts, sending it around the circuit through the resistor (jagged segments on the right) and back. When V is high, the capacitor gets charged more, and then it gives back some charge when V drops, smoothing the current through the resistor.
Series, parallel, and series-parallel combinations: Your instructor
and/or your text will derive
Cs = 1/S C and Cp
= S C. It is important to know that
series capacitors have the same charge on them, and parallel capacitors have
the same voltage across them. The typical problem is to find the charge of, and voltage across, each capacitor.
The circuit below has a 12 V battery on the left. The capacitors are 3, 2, 6, and 12 m f. Initially the switch is open, so only the 3 and 2 m f capacitors are charged. Find q and V of each. b)Then the switch is closed. Now find q and V of each.
Answers:
a) The 3 and 2 m f each have 14.4 m C of charge on them; the 3 has 4.8 V across
it and the 2 m f has 7.2 V across it.
b) 3 m f cap
has 24 m C and 8 volts across it.
2 m f has 8 m C and 4 volts across it.
6 m f has 16 m C and 8/3 volts across it.
12 m f: has 16 m C and 4/3 volts across it.
An efficient way of doing these problems is to first find the capacitance of the whole circuit. With the switch open, we have only the 3 and 2 m f in series. By the way, the series formula for two capacitors reduces to the product divided by the sum, or 6/5 m f in this case. In series, each must have the same charge…. You figure out the rest of it. When the switch is closed, the 6-12 series combination is in parallel with the 2, and that whole mess is in series with the 3. See if you can come up with a total capacitance of 2 m f, then find the charge and voltage on the 3 m f….
Imagine traveling clockwise through the circuit starting on the lower left. When you go across the battery, your voltage is raised up 12 volts. Step across the 3 m f and then the 2 m f, and your voltage must drop down a total of 12 volts. Or go the long way….
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