The the most prevalent analogy for understanding electricity is the water hose one. I personally like this because it's very understandable but I have a question about how this works with resistors.

I understand that resistors will limit current which because of ohms law will affect the voltage but in a hose, decreasing the hose diameter (current) would increase the water pressure (voltage). Am I wrong on my understanding of this principle?

  • \$\begingroup\$ current is not the hose diameter, it's how much water flows per second. \$\endgroup\$
    – 1p2r3k4t
    Oct 15, 2013 at 14:42
  • \$\begingroup\$ Let us assume inlet of hose is connected with the over head tank (voltage source) and the outlet of hose is connected with the ground tank.we know water will flow from higher pressure to lower pressure point;so water will flow from OHT to ground tank(if the pressure at the outlet of hose is also same as that in the inlet of hose there wont be any water flow) \$\endgroup\$
    – yogece
    Oct 15, 2013 at 14:43
  • \$\begingroup\$ Water as an analogy for resistive electricity, like most analogies, an imperfect comparison for the sake of teaching. \$\endgroup\$
    – Passerby
    Oct 15, 2013 at 15:30

4 Answers 4


I'm guessing this is the phenomena you are imagining:

You hold the end of a garden hose with nothing attached, and turn on the spigot. Some water gushes out, with high volume, but low velocity. Then, you restrict the end of the hose with your thumb. The volume of the water is less, but the pressure in the hose rises, and it shoots much farther.

Hydraulic systems can make good analogies for electric systems. The problem is that not all hydraulic systems have an electric analog, and some people suck at making hydraulic analogies.

Here's one problem with mapping this hydraulic system to an electric system: charge is conserved, both in the universe as a whole, and practically speaking, most circuits. With your garden hose, an unlimited supply of water is magically created by the city water system, and when it exits the hose and hits the ground, it's no longer in the circuit.

To get the electric analog of this, you need a circuit capable of shooting charged particles out into space. You also need something capable of supplying charged particles. Things like this do exist (for example, a CRT flings electrons through a near vacuum at the phosphor coating on the screen) but they typically require high voltages, and you aren't going to build anything like this with just a resistor and a battery. If you connect a resistor across a battery, the electric charge is pumped around the circuit. No charge enters, and no charge leaves.

Another problem: very commonly, electric circuits are designed to maintain a constant voltage. The water supply system is also designed to maintain a (roughly) constant pressure. However, since nothing (electric or hydraulic) can supply unlimited current, all of these voltage/pressure regulation systems have limits. In the case of your garden hose, the water supply system can't supply enough water to keep the pressure at the end of the hose at the target, say 30 psi. With nothing attached to the hose, there isn't enough resistance for the supply to work against to build the pressure. It is analogous to an electric short circuit.

If you were to block the hose entirely, you would find the pressure inside the hose (and indeed, everywhere in your hydraulic system, if the hight differences are irrelevant) will be 30 psi. If you were to open the end just a little bit, you'd find that the pressure is still pretty much 30 psi. Only until quite a lot of water is flowing would the pressure drop from 30 psi; this is because at high currents, the friction of the water flowing in the hose becomes significant, and increasingly more of the pressure is lost over the hose's resistance as the current increases.

If we wanted to model the hose-thumb system electrically, we'd need to take into account that the hose has some friction. Maybe something like this:


simulate this circuit – Schematic created using CircuitLab

This is called a voltage divider. When your finger is not on the hose, (\$0\Omega\$), the \$5\Omega\$ from the hose is very significant, since it's the biggest resistance in the system. When your finger is blocking most of the hose (let's say \$1000\Omega\$), then the additional \$5\Omega\$ from the hose makes very little difference.

but in a hose, decreasing the hose diameter (current) would increase the water pressure (voltage).

So with that explained, we can circle back to your confusion. It depends on where you measure the pressure. You probably know that if you totally block the hose, the water pressure does not rise without bound (your pipes would burst!). Using thinner or longer pipes actually decreases the water pressure available at the spigot (or appliance, or whatever is connected), because more pressure is lost to friction between the supply and the spigot. There's an electric analog: higher currents require fatter wire.


In the water analogy voltage is presented by water level, not pressure (though there's a relationship between the two). A 12 V across the resistor is equivalent to a (for instance) 12 m water level difference between the ends of the hose. If there's no water level difference ( no voltage difference) then there's no current, no matter what the pressure is.

  • \$\begingroup\$ So using this analogy, the current could be slowed down... (the flow) but the voltage could remain the same if it was a constant supply (the water level)? But there is also voltage consumed by the resistor correct? \$\endgroup\$
    – G Thompson
    Oct 15, 2013 at 14:41
  • \$\begingroup\$ There is more than one water analogy. \$\endgroup\$
    – Phil Frost
    Oct 15, 2013 at 14:43
  • \$\begingroup\$ @GThompson: no, voltage is not consumed, but current is. The voltage is across the resistor, but doesn't necessarily change, while there's a current flowing. The water flow represents that consumption, and if the power source is a capacitor the flow will decrease the voltage difference, just like water flowing out of a tank will decrease the water level there. \$\endgroup\$
    – Johan.A
    Oct 15, 2013 at 14:43
  • \$\begingroup\$ @Johan.A I appreciate it. So say I have a 12volt power supply and I have a 5volt DC motor. I can use resitors to bring the voltage down enough to keep it safe correct? Would I need to know the resistance of the Motor itself? \$\endgroup\$
    – G Thompson
    Oct 15, 2013 at 14:54
  • \$\begingroup\$ Not really, no; the bad idea of using resistor dividers for this purpose keeps coming up and is still a bad idea. \$\endgroup\$
    – pjc50
    Oct 15, 2013 at 15:03

Every analogy can be dangerous because is difficult to consider all things. It helps you understand, but you cannot consider it as a perfect model. Resistor will limit current considering you have a fixed source voltage. The ideal source voltage will not variate.


In the water analogy, a resistor is a constriction in a pipe. If you externally always force a certain flow rate (current) thru the pipe, then yes, the pressure will increase. Real resistors work that way too. If you always force a certain current thru a resistor, the voltage (pressure) will go up if the resistor goes up.

In the case of a fixed pressure (voltage) accross the ends of the pipe, the flow rate (current) will go down as the constriction gets more narrow (resistance goes up), just like it works with real resistors too.


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