Among other things I don’t understand, I don’t understand how electrical resistance works, other than it’s important enough that it has its own measurement, ohms. My electric range top works because the coils resist the flow of electricity and turn it into heat. OK, so why does an insulator block electricity altogether, rather than turn it into heat? Is an insulator not the same thing as a very high resistor?
Resistors are elements which conduct current, albeit with a drop in voltage from one end to the other. They’re made of conductors of some sort, which allow the passage of current (although possibly a little reluctantly), and can be made of metal (although metal doesn’t allow much of a voltage drop, which is often the point), films of carbon or other material (probably the most common sort of hand-holdable), tubes filled with salt solutions (I’ve used these), or other things.
Resistors:
Conductor:
Insulator:
You might want to get a good introductory book like this:
(No offense intended. I owe stacks of “X for Dummies” books.)
Yes. An insulator is a resistor with a very high resistance.
Materials that conduct electricity such as metals allow the free movement of electrons from atom to atom or molecule to molecule. Atoms of metals don’t do a very good job of holding on to their outer shell of electrons and they move around conducting electricity when a charge is introduced.
An insulator does a very good job of holding onto those electrons. Introduce a charge to an insulator and it sits there and does nothing. The outer electrons do not jump ship.
A resistance like the rings on your stove, heats up because the electrons are getting ‘ticked off’ at not being able to move through easily - the actual scientific term is Joule Heating - the electrons bump into other ions and give up some of their kinetic energy as heat. The key here is that the electrons DO move through the resistance. In an insulator, they don’t move through the material so the material doesn’t heat up.
Make sense?
A possible analogy is to consider the flow of electricity to act like the flow of water. Imagine that water held in place by a solid concrete damn. It has resistance to the flow of the water and in effect infinite resistance. That damn is an insulator.
Now imagine that water held in place by a dam made of bales of hay. The hay would have the ability to hold back the water for a while but water would seep through at a fairly quick rate. The hay dam is a resistor.
Imagine a dam made of chicken wire. The chicken wire has some resistance to the water but very little.
The electric insulators and resistors that you noted act the way they do because their properties either totally or partially impede the flow of electrons. The materials that make up those objects determine how electrically “porous” those objects behave.
Everything’s a conductor, given enough voltage. And everything’s an insulator, given not enough. 
In order to get electricity to flow any material, you need to provide enough of a “push” to overcome the resistance of the material. This process always produces heat; the higher the resistance, the more heat is produced.
In the case of the rubber surrounding a wire, the rubber’s resistance is so high that the expected electric potential in the wire would never break through it. On the other hand, a lightning strike has such massively gigantic voltage that it can be conducted by a tree, even though wood is normally thought of as an insulator.
Just a bit of fun, since the OP has been answered.
A capacitor is a couple of conducting plates with an insulated gap between them.
Anyone care to explain to the OP how current flows through a capacitor?
You are absolutely right, it is.
The first thing to understand is that a high resistance means very little electric current flows. For an insulator, the resistance is high enough that the amount of current flowing is so small that we can consider it zero. And of course, the opposite is true: with low resistance, there’s more current flowing (that’s why we call things with very low resistance “conductors”).
The second thing to understand is that the heat given off depends on how much current is flowing (makes sense, right? More electricity flowing means more stuff going on and more heat given off, and if electricity isn’t flowing at all, there’s no heat at all).
Now, it turns out that with a power source like a wall socket, you end up with the most heat by plugging in the lowest resistance you can find, so you get as much electricity flowing as possible. That’s why short circuits cause fires: there’s tons of heat being given off.
To make an electric stove, you actually have to have a medium resistance: low enough that you get enough current flowing to get heat, but high enough that nothing gets so hot that it melts.
[I see others got in while I was writing this, but I’ll leave it here, since it can’t hurt.]
Yes, an insulator is a very high resistance resistor. It’s things with low resistance that heat up, not things with high resistance.
There are two key equations, here. First, the definition of resistance: R = V/I (this is often mistakenly called Ohm’s Law). Resistance of a circuit element is defined as the ratio of the voltage across the element to the current through the element. So if I put a voltage across something with a very high resistance (like, say, rubber), I’ll get a very low current through it. But if I put a voltage across something with a very low resistance (like a copper wire), I’ll get a very high current through it.
Second, there’s the equation for the energy lost to heat. The power of a circuit element is equal to the current going through it times the voltage across it: P = IV. This energy can in principle be in many forms: In an electric motor, the electrical energy is being turned into kinetic energy, and in a light bulb, it’s being turned into light. So, for instance, a 60 watt light bulb plugged into a 120 volt socket must be drawing 0.5 amps, and a 1 horsepower (=750 watts) motor plugged into a 120 volt outlet must be drawing 6.25 amps. For a resistor, though (and everything is a resistor, to at least some degree), that’s all energy that’s being turned straight into heat.
Now, there are many ways you could in principle build a power supply, but in practical terms, most of the power supplies we use (batteries, the power from the electric utility, etc.) are designed such that they maintain a constant voltage, no matter what (if anything) they’re hooked up to. This means that if you hook your power supply up to things with different resistances, you will necessarily have different currents. If I press a nine-volt battery into a block of rubber, my circuit has extremely high resistance, so an extremely low current will flow, even though the battery is still at 9 volts. 9 volts times an extremely low current is an extremely low power, so the block of clay will heat up only by a negligible amount. On the other hand, I could press a nine-volt battery into a pad of steel wool, which (like any metal) has a very low resistance. Since it has a low resistance, a very high current will flow, and a high current times 9 volts is a high power, so a lot of energy is converted to heat, and the steel wool will burst into flames.
Incidentally, people often use “short circuit” as a general term for “something’s not working”, but it actually refers to a specific sort of fault. A short circuit is what happens when something with very low resistance (a bit of stray wire, say) connects two points that are at very different voltage (such as the terminals of a wall socket), causing a very high current to flow. This causes the wires in the circuit to heat up, and could (if not stopped) start a fire. The purpose of a fuse or circuit breaker is to blow if the current gets too high, to break the circuit and prevent more expensive things from burning out (like your house).
Small Clanger, current does not flow through a[n ideal] capacitor. Electric potential builds on one plate, which forces a depletion on the other plate, so current can flow in the circuit but in principle nothing is flowing through the dielectrc.
Just want to follow up on this. Once the capacitor has been fully discharged and the circuit comes to a steady state, there will be no current flowing anywhere through the circuit.
As mentioned, DC doesn’t flow thru a cap. There is a momentary charge build up on one side which causes a charge on the other side as voltage is applied. AC can ~flow thru by alternate charging and discharging. Handy for allowing a signal across it but blocking any bias voltage. (And because of the delay, they can also be used to smooth out currents.)
In the water and pipe analogy, think of it as a rubber membrane blocking a pipe. The water itself cannot flow, but any oscillation in pressure gets transferred across.
Depends how you define DC.
If you define DC as the steady-state, mean value of a waveform (a.k.a. “DC component” and “DC bias”), then you’re correct. But if you define DC as “current flows in one direction only” - which is the classic definition of DC - then DC can flow through a capacitor. If you want proof of this, charge a capacitor using a 9 V battery. Current flowed in one direction during the charging process, hence DC flowed through the capacitor. Using the classic definition, DC can indeed flow through a capacitor as long as its magnitude is varying over time.
Having said that, when someone says DC can’t flow through a capacitor, an assumption is that they’re defining DC as the steady-state, mean value of a waveform, and thus their statement would be correct.
This is generally true in practice, but only because constant voltage sources - including things that can be modeled as such - are much more common than constant current sources. If constant current sources were vastly more common than constant voltage sources, then we would say the opposite.
True, but it should also be mentioned that EEs only like to talk about absolute resistance when referring to a device where voltage is (more or less) linear with current. An example where this is not the case… you can easily measure the voltage across a diode and the current through it, and then calculate its absolute resistance (R = V/I), but EEs would laugh at you if you did this. If you said, for example, “The voltage across this diode is 0.62 V and the current through it is 0.25 A, therefore its resistance is 2.48 ohms,” the EEs would be snickering. That’s because the voltage-current relationship of a diode is not linear, hence talking about its absolute resistance is pretty much meaningless. Generally speaking, the only time we talk about the “resistance” of a nonlinear device is when we’re talking about its small-signal resistance, which is something else altogether.
Power=current squared times resistance.
Current= voltage devided by resistance.
With a low resistance the current will be high, now square that to get the power (heat) given off.
Increase the resistance by 1,000,000 times and the current will be extreamly low, square that and you have a much smaller number to times the resistance a lot less heat.
example:
V=208 volts
R=1 ohm
then I would =208 amps, and power would be 208X208X1=43264 watts of power.
V=208
R=1,000,000 ohms
then I would =.000208 amps, and power would be (.000208)X(.000208)X1000000=.043264 watts.
Does that shhow why insulators do not get as hot.
That’s pretty much it. It also shows why power companies step up the voltages so high. If everything were transmitted at household voltages the current would be extremely high and the I^2*R losses on the lines would be unmanageable.
No, that just shows that current is flowing through other circuit elements which are in series with the capacitor. Current flows into one end of the capacitor, and it flows out of the other end, but it never actually flows through it.
An even simpler example where it’s not the case is any circuit element which heats up significantly, such as a light bulb. Though there one might be more inclined to say that the resistance varies with temperature, rather than that the element doesn’t have a well-defined resistance.
You’re correct that charges do not physically flow through the dielectric. But this bit of low-level detail usually doesn’t matter when analyzing circuits. I just pretend that charges flow through the capacitor under certain conditions, and everything works out O.K. So while it is technically incorrect, I always say “current through the capacitor” just to keep things simple.
True. But unlike a diode, a light bulb is either completely off or completely on (i.e. operated at 100% of rated voltage) in most application. So I don’t have a problem with someone talking about the absolute resistance of a light bulb, but they must also tell me if the resistance value is for the off state or the fully-on state.
Snnipe 70E:
Your analysis is correct, but we must be careful not to conclude from it that the lower the load resistance is, the higher the power dissipated by the load. At least for real applications.
I am assuming the 208 V source in your example is grid power. If this voltage source was ideal, then indeed, the lower the load resistance is, the higher the power dissipated by the load. But real grid power has a built in source resistance. The load power will be at a maximum when it equals the source resistance of the voltage source. If, for example, the 208 V grid power had a source resistance of 0.01 ohms, the load resistance will dissipate maximum power when its resistance is 0.01 ohms. As the load resistance goes higher than 0.01 ohms, the power dissipated by the load will decrease. As the load resistance goes lower than 0.01 ohms, the power dissipated by the load will decrease. If the load resistance = 0 ohms, the power dissipated at the load will be 0 W. Just something to keep in mind.
Of course, that means that all the power (excepting transmission losses) will be dissipated at the source, which tends to piss off the power company. So, don’t do that. 
But it’s so much fun!
All of this analysis is “first order,” of course. Under real-world conditions things can get complicated in a hurry. Unless you’re talking about a superconductor, there’s no such thing as a zero-ohm load. If you do create a short, and if the CB does not trip, the temperature of the conductors will rise very rapidly. But we must not forget that metallic conductors have a positive temperature coefficient of resistance, which tends to decrease the dissipated power. And then there’s the contact resistance at the short, the plasma in the arc, rapid oxidation, blaa blaa blaa.
Not when you’re standing right next to a phase-to-phase fault at 480 V and your hand is holding the meter probe which dislodged the improperly fastened wire which caused it.