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I'm currently getting started to learn about electrical engineering, and I'm having some difficulty separating voltage from current.

I have seen a couple of times that a battery is compared to two water tanks, one filled with lots of water, and the other one empty. Between them is a hose that connects them.

Now there is some amount of water that flows through the hose in a given amount of time, e.g. 1 liter per minute. Is this equivalent to current? In other words: Is the amount of electrons that are "transferred" in a given amount of time measured in Ampere? Is this right? So, again in other words: Is current basically comparable to something such as kilometers or miles per hour, so is current the speed of the electricity?

As far as I understood voltage is the difference between the amount of water in the two tanks. So it's like the pressure. In the beginning, when one tank is full, and the other one is empty, the pressure is higher, than after some time, when some water has already flown. Does this mean that the voltage of a 9V battery is only 9V in the beginning of its lifetime, and gets less over time?

Now I wonder about their relationship: If the voltage is higher, it does not necessarily mean that the electrons flow faster, only that there is more pressure behind them to get over resistances. Right?

And, when we are now talking about resistance: A resistance is comparable to someone who steps onto the hose and lowers the amount of water that can flow. Transferred, does this mean that a resistance lowers the current, but not the voltage?

Sorry if all these questions seem to be dumb, but as I said, I'm really at the very beginnings, and I'm trying to get the concepts right.

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    \$\begingroup\$ The problem with any water analogy of electricity is that some parts of it will be incorrect and other parts will be insufficient. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 28 '15 at 23:34
  • \$\begingroup\$ Okay, that's basically the law of leaky abstractions, but what exactly did I get right, and what did I get wrong? \$\endgroup\$ – Golo Roden Feb 28 '15 at 23:35
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    \$\begingroup\$ Ampere = "One Coulomb of electrons passing a point per second", Volt = Pressure needed to move ampere through a resistance, Ohm = Volts/Ampere, Watt = Volt times Amperes in DC circuits. \$\endgroup\$ – Optionparty Mar 1 '15 at 0:56
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    \$\begingroup\$ I'm with @IgnacioVazquez-Abrams. I almost "hate" water analogy for electrical circuits. I've seen too many students grow too fond of it and become completely confused when trying to understand a little more advanced circuit than the most basic ones. The problem, IMO, is that the equations of the fluid mechanics and those of electromagnetism are quite different, so the analogy can be only superficial and is beneficial only in the most trivial cases. In my experience, water analogy will get in the way when you advance in your studies, and make things more obscure. ... \$\endgroup\$ – Lorenzo Donati supports Monica Mar 1 '15 at 1:47
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    \$\begingroup\$ I see one issue: current is not the speed of the electrons, but the rate of flow. Low current is a car on the highway moving 100 km/h; high current is a whole line of cars moving together. \$\endgroup\$ – Greg d'Eon Mar 1 '15 at 5:01
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Basically. Your analogies are right, but unfortunately they don't work for everything.

Current is the amount of electrons passing through one point per second. The more resistance you have the more voltage is required to get the same current.

Some fundamental equations for DC circuits are:

  • V=IR (voltage = current * resistance)
  • P=IV (power (watts) = current * voltage)
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  • \$\begingroup\$ Thanks for your answer. It helped me understand Ohm's law :-). Basically V=I*R just means: If you want to get 1A over a resistance of 1 Ohm, you need 1V. For each Ohm you add, you need one more Volt. And if you want to increase the Ampere, you also need to increase the voltage. Thanks, I think I got it :-) \$\endgroup\$ – Golo Roden Mar 1 '15 at 21:18
  • \$\begingroup\$ Yep. Glad I could help! \$\endgroup\$ – Reid Mar 1 '15 at 23:15
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Voltage is analogous to pressure. Water flow rate is analogous to current. Drag (due to narrow pipe or long pipe) is analogous to resistance. If you reduce the flow rate in a hose by stepping on it, you will also affect the pressure. Basically, you will increase the pressure drop across the restricted area. Likewise, if you add a resistor in series with a wire, you will affect the voltage on one or both sides of the wire as well as the current. Basically, you will increase the voltage difference across the area where the resistor was added.

For DC circuits, you can think of it this way: electrons entering a wire on one side push out electrons on the other side just like water molecules entering a pipe. I guess the difference is that copper pipe is always full of electrons. They can't leak out (practically speaking).

If you need to understand what happens when voltage or pressure changes rapidly, then look at it like this: pressure disturbances travel at the speed of sound, and voltage disturbances travel at the speed of light (the speed of light as measured in the dielectric material where the wave is traveling).

I don't find that there is any big problem with water analogies, but I think many people don't really understand how water flow works, so they carry that confusion over to electric circuits as well.

Anyway, it is fine to try to gain intuitive understanding. But you should try to internalize Ohm's law and make that your new intuition: V=IR

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  • \$\begingroup\$ Your last sentence was the most helpful, and in combination with @Reid's answer, I think I got an idea that works for me. Thanks for your help :-) \$\endgroup\$ – Golo Roden Mar 1 '15 at 21:20

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