Lindy Your summary sounds like what was said by various people but has altered subtly, by enough to confuse you. The analogy is simple but MUST be applied correctly to help you. As you have interpreted it its actually wrong. It only needs a tweak, but without the tweak it will mislead and confuse you. There is close to a 1:1 mapping for of terms for a simple use of the analogy. NB - (NBNBNB) in the following I have NOT said what "water" is. Current will be seen to be the RATE of water flow. Volts will be seen to be the pumping pressure. Resistance will be seen to be the difficulty that a given amount of pressure has in causing flow to happen. BUT I do not say what the *water* actually *is* in electrical terms. Let it be so for now. None of the basic descriptions do either - but nobody usually notices. For now just think of water being pumped. [[[Pssst: "charge". Ssssh. Not yet]]] All we are saying so far is V =3D I x R Volts =3D Current x Resistance And the rearrangements of this. Add or change and the model presented so far fails. _____________ In the following "time" attempts to get introduced. Do not let it, yet. Time does not feature in the basic equations so far. Adding time will introduce us to energy. We have not got to energy yet ____________ > Here is where I learned something new. > Amps isn't the water pressure, True. > it's how much water pressure that can be delivered depending on how much is necessary. No. Amps are the RATE of water flow. (Gallons per second say)(=3D volume per time. In our analog its volume of water. In the electrical model we do not say what it is volume of. Not yet.). Amps are NOT water-pressure and Amps are NOT water-volume (ie Volume per time is not volume, it's rate. You have to know time to get volume and we have not talked about time yet). Knowing the current flow does NOT tell you the pressure. (Delivering water-pressure may have a meaning but its not what is being said here). If you monitor over time then you would be able to tell the volume that had flowed but without time an instantaneous measure. Volts =3D pressure =3D pumping head =3D height of a dam providing pressure. An older term for Volts is EMF =3D Electromotive Force. > In other words, I won't know how much current is available until I need it. Sort of. You don't know what the rate of flow will be (Amps) until you know the pressure applied (Volts) and the resistance to flow in your lod (resistance= ) > A cat's whisker across a 9 volt battery will melt (like the igniters in a model rocket engine). Very likely, BUT you have introduced two (at least) new concepts which are related but will hopelessly confuse you if you do not 1st adequately understand the basics. You have added energy dissipation and you have added temperature rise (and melting). These are related BUT until you know the characteristics of the cats whisker you cannot know how much energy will be dissipated or the temperature rise or the result. Your conclusions are PROBABLY correct but we have got into "somewhat like" rather than "almost exact analogy". Lets get the analogy right first. > But a huge copper wire across the poles of my imaginary battery would melt it completely because of the potential. I don't know if "it" means battery or wire will melt. Same comments apply. If the huge copper wire has very low resistance (as it will have) and if the super-battery is "ideal" and has no resistance then yes, very large currents will flow , very large energy will be dissipated and the copper will probably melt. But the same comments on new terms applies. > If I follow the water analogy, I can open the dam all I want, but if there is no water to back it up I get nothing downstream. This statement makes sense but has departed majorly from the analogy. The dam mentioned is meant ONLY to be a full dam and represents pressure by the height of water in it. Water head =3D pressure =3D Volts. If the dam is empty there is no pressure. "Opening the dam" is an understandable term but leads us to switches which is not what the analogy intended. Safest is voltage =3D =3D pressure. The dam picture was meant to help by adding water-head as an analogy to pressure but should not be extended to other aspects of dams. Volts =3D water pressure (symbol =3D V) Amps =3D rate of flow (symbol =3D I) Resistance =3D Volts required to cause a flow rate of 1 Amp. ie how much th= e pipe or whatever resists the applied voltage. (Symbol =3D R) The following formula is the same one arranged in 3 ways. UNDERSTAND and remember any one version and you have all 3. V =3D I x R R =3D V / I I =3D V / R Also - and not covered yet is Power=3DV x I (Units W =3D Watts) W =3D V x I Substituting from above also gives W =3D I^2 x R W =3D V^2 / R These are the same formula - just rearranged and using the relationships fro V =3D I x R __________ > Then volts is how fast it flows because of the incline of the river. So, No. Current is how fast "it" flows. Volts ARE the incline. We haven't got to what "it" is yet. > Dagmammit, is this a difficult concept for others, or just me? > It's difficult, as is anything new, until it becomes easy :-). Riding a bicycle is an impossibly hard task, until you can ride one. Flying a helicopter is beyond even the very top jet pilots. Until they can fly one. Walking is beyond the very most capable of babies. But someday some of them will run sub 10 second 100 metres. But you MUST stick doggedly to the analogies provided until they make sense and/or burn in. If you tweak the analogies it will never work. Russell --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .