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Since Voltage is like water pressure, and Amperage is like the volume of water being pumped, in a given hose (wire), there's a volume threshold based upon its diameter (AWG), but only in relation to the pressure applied (V).

So, if I pour water into a closed end hose, it will fill up and spill out. There is very little or no pressure at all. Just gravity.

If I add pressure, I create friction, the hose expands by resisting the pressure (light slowly fades in). Of course, it's a hose, and it has water in it, and therefore it probably won't catch on fire from friction. If I add too much pressure for the hose (AWG) it will burst. If it was a wire, it would heat up and turn into a light saber, which would be great for an oven, or a foam cutter, or a distant galaxy far, far away.

Am I on the right track?

So, the relationship between applied current and burning your house down correlates to the thickness of your wire (whether it's solid or stranded, what material it's made of, etc.,) and the amount of current you apply to that wire.

QUESTION PART:

Where I'm confused is how amperage works. In a hose, or a water line, for example, each outlet valve reduces the volume of water being sent to the next valve until the pressure is reduced to nothing. If we want the pressure to remain constant, we would need to reduce the pipe diameter as the line continues.

In electricity, it seems like it's similar, only the electrical devices draw amperage...yes? Wouldn't that be like adding a pump to the main water line to pump already pressurized water out of the pipe, in which case, if we did, we may deplete the water line too quickly, causing it to collapse...

Is that what's happening when we fire up too many devices that draw more amperage than the breaker can handle?

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    \$\begingroup\$ It's not a good analogy. Everyone uses it. And it's just as bad as it always will be. Water is a liquid. Liquids do not behave as a gas does and it's better to treat electrons as the Drude model does, as a gas. So your analogy needs to be re-cast more as an air compressor system. \$\endgroup\$ – jonk Dec 21 '19 at 18:24
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    \$\begingroup\$ The device doesn't draw amperage, it allows amperage. A device (non-inductive) is essentially a resistor across the voltage terminals. Its internal resistance allows the voltage to push itself through the load. The resulting current flow is determined by Ohm's Law. This simple example uses an unchanging DC source and a purely resistive load. \$\endgroup\$ – John Canon Dec 21 '19 at 18:43
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    \$\begingroup\$ More specific to your question, the electric charge of an electrical circuit and its elements is constant, to a first approximation. There is no equivalent for a liquid system using water, where the amount of flowing water isn't constant in the situation you just pointed out. So the water analogy fails miserably here. In an air compressor situation where the system pressure is maintained by a generator/compressor that monitors it, this problem goes away and therefore the air analogy has broader application and is, I think, much improved. It just isn't offered often. Too bad. \$\endgroup\$ – jonk Dec 21 '19 at 18:44
  • \$\begingroup\$ In small towns they use large reservoirs above ground to maintain constant water pressure by gravity then use massive pumps to keep it full. The size reduction causes eddy current flow resistance just as it does on the grid with AC skin effects. The wire resistance & inductance per km are well defined and is usually chosen to allow no more than 10% voltage variation but use tap changers in large >10MVA units to regulate a feed .The body uses chemical protein methods to modulate artery and vein diameter to regulate pressure and flow. Flow restriction & pressure turns red.like heater wire . \$\endgroup\$ – Tony Stewart EE75 Dec 21 '19 at 18:55
  • \$\begingroup\$ However closing a switch would have to perform like opening a valve to flow , so the analogy fails. The resistance of the source must be 10 times lower than a heater feed wire and heater wire in order to keep the voltage from dropping more than 10% which you may have noticed in your past experience. \$\endgroup\$ – Tony Stewart EE75 Dec 21 '19 at 19:08
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Don't forget: the water analogy requires closed loops of pipe. No water pours out anywhere. No pipes change their diameter (no inflation or deflation.) It's not like a long cup of water. It's more like a rotating wheel.

An electric circuit is a ring of pipe, full of water, with no bubbles allowed. Next, add a constriction, and that's a resistor. Add a water-pump into the ring, and that's a power supply. The speed of the water is proportional to amperes, with fast rotation being high current. Voltage is the pressure-difference found across the resistor (or across the pump.) DC is when the water moves continuously. AC is when the water wiggles back and forth (yet notice that any resistor heats up, regardless of the flow-direction.)

Parallel circuits are when we add a couple of "T" junctions to our water-ring, so the path splits at one point, then recombines at another.

A similar electricity analogy is the bicycle wheel, where the rubber of the tire becomes the electricity found inside a closed circle of wire. Push the tire with your hand, so that it spins, and that's the power supply causing a current. Rub the spinning tire with your thumb, and that's a resistor. Mechanical energy flows almost instantly, going from your pushing-hand to your rubbing-thumb. The current starts everywhere, all at once. But the path of current is a closed circle, with no rubber being consumed. And, the electrons (the rubber) move slowly, moving like a wheel.

Electricity isn't a "form of energy," any more than the rubber of a tire is a "form of energy." The electricity (and the rubber) were already there, ever since the circuit was created. And, whenever electricity flows, it goes in a complete circle, with none being gained or lost. Your hand doesn't generate electricity, instead it just causes the existing electricity to start moving. (And in the water-hoses, the pump doesn't generate any electricity. It just takes electricity in through one terminal, while spitting electricity out the other terminal at the same time.)

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There are a couple details that you should think through in more detail.

First, the capped off pipe really is a good analogy to an unconnected wire: Water & charge flows until the far end has enough accumulated water / charge to push back via pressure / electric field.

Second: if there are no frictional pipe losses / resistance, then adding new connections just results in more water / charge flow. There are no pressure / voltage losses in this case. So long as the source can keep up, more draw just draws more.

When there is frictional pipe losses / resistance, you get to the interesting case. As you pull more water / current through, there are more losses in the feeder pipe / wire: pressure / voltage drops. With water, those losses don’t generally heat up the large mass of water & pipe. But those losses in the electrical case can be significant: they can heat up the wire to the point that it can cause a fire.

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