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Ohm's Law: Electricity vs. Water

posted Sep 30, 2012, 6:16 PM by Charles Boling
No, we're not electrocuting people in the bathtub.  KF7CVT and I were discussing the use of the classic analogy of water to electricity when explaining the concepts between Ohm's law.

This is an extract from that discussion, with a few comparisons that illustrate similarities in flow control.  It's not top-quality prose -- a bit like reading someone's classroom notes -- but perhaps it will turn a light on over someone's head.  The electricity might, if the water doesn't! ;-) 

E(lectricity): Start with a battery, say 12V.  With no load across it, 
you can measure 12V across the terminals. W(ater): Your fresh water system is set to a max pressure of, say, 60psi. That's what the gauge shows with no faucet running. E: Place a small load across the battery, say a light bulb. You won't see any [measurable] voltage drop across the battery terminals, because the battery's able to handle the load fine. Same if you measure across the light bulb's terminals, because the wires have almost zero resistance. Stick an ammeter in series with the bulb, and you'll measure the amount of current that the bulb's resistance allows to pass. (The bulb acts to restrict the flow of electrons). W: Open a faucet a little. Your pump will still keep your pressure @ 60PSI (or close to it) because it can handle the flow just fine. if you stuck your finger right against the water coming out of the valve, you'd feel it hitting you at 60PSI pressure, over a small area (because the valve's only open a little presenting a small hole where the water comes out). E: You could add a resistor in series with the light bulb to limit the current flow. Even if you don't, a real-life battery can't provide an infinite amount of current -- it has a significant "internal resistance". W: You could put a valve (or smaller pipe) upstream from the faucet to restrict the amount of water available to the faucet. Even if you don't, your pump can only put out a finite flow -- its parts are only so big and can only go so fast. E: If you try to put more of a load on the battery than it can handle, you'll start to see the voltage across the battery drop. The bigger the load, the more the voltage will drop. Short it out, and the potential drops to zero. W: If you open the firehose you have attached to your residential pump, you'll see your pressure drop. It won't shoot out the hose very far, and the person who was in the shower at the time will complain that their flow suddenly went to a trickle. More water comes out of your hose than the shower because it's the easier path, but the pressure at the end of your huge wide-open hose isn't great. Put a nozzle on it and start restricting the flow, and the pressure will start to rise, making it shoot out further, but not providing as much water flow to douse a fire with. E: Picture the internal resistance of the battery as a resister attached to it, placed in series with the rest of your load. Whether the "resistor" is an external component or part of the battery chemistry/structure, it doesn't matter. W: Picture the physical current limitation of the pump as a pipe section of a particular diameter. Doesn't really matter if the smallest section of pipe is part of the pump or downstream a ways, it's still there. E: Why does the voltage drop with a heavy load, and can it be calculated? Remember the "built-in resister"? That and your load together effectively form a voltage divider, as shown here:

Voltage divider using two resisters

Consider the internal resistance of the battery (and resistance of the wires themselves) as R1, and your load as R2. When you measure the voltage across the bulb, you're measuring the voltage drop of the divider. Voltage dividers are very simple. If R1 = R2, then the voltage across R2 is 1/2 the original voltage. If, say, R1 = 3 ohms and R2 = 9, then the voltage across R2 will be 75% of 12, or 9V. If R1=75 and R2=25, the the voltage across R2 will be only 25% of 22, or 3V. If you independently measure the resistance of R2 (your load) and see what the voltage drop is when connected to your battery, you can use Ohm's law to calculate the internal resistance of the battery (R1) -- we'll assume the resistance of the wires is negligible, in this case. W: Similarly, by measuring the pressure drop as the valve is open, you can figure out about what the effective diameter of the pump opening is. E: Real-world power supplies always have some resistance, which is why you'll never get infinite current if you short them out. W: Real-world water sources always have restictions; you can't fill the ocean in 5 seconds. E: Measuring directly across the power source -- across R1 & R2 together (we'll now assume now that R1 is a resister in series with your load, rather than the being the internal resistance of the battery, so we *can* get our probe across it), you'll still have the same 12V. It's only when you put your probe past R1 -- at "Vout" -- that you see the voltage drop. Note that restricting the flow of current (or totally shutting it off by breaking the wire) will never rase the voltage above 12V. W: The pressure before the 1st restricting section will measure 60PSI, but past the restriction, where you open again into a larger point, the pressure will drop. However, if you shut the end of the hose off, once the hose fills, the pressure in all parts will build back up to 60PSI. At no place will you ever see more than 60PSI. Well, that thing with internal resistance is probably more than you wanted for simple Ohms law calcs, but it's real-world, and explains why real-world batteries and hoses behave the way they do. Moving on to power (V*I=P): W: Cheap pressure washers scream "30 million PSI of cleaning power" E: Spark plug coils (or home-made shockers) scream "20,000 volts!" W: Only if you're putting out a needle-size stream flow of a pint a minute! E: Yeah, at .00000000001A... W: Not going to clean your deck very fast... E: Not going to electrocute your friend, or move that motor far... W: Columbia river has WAY more flow than the pressure washer... E: 30 million dry cells in parallel has lots of available current W: 0.1 PSI isn't going to blast dirt off of anything E: 1.5V isn't going to be able to push that available current through much to do useful work W: Pressure washers rated in cleaning units -- GPM * PSI E: Motors rated in Watts -- Amps * Volts W: Minimum pressure needed to loosen dirt; once that's satisfied, but greater total power gets the job done faster
E: Yeah, kind of the the same... There are reasons your industrial plant doesn't run on 1.5V...