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This is a pretty basic question, but I'm new to electronics, and I'm a little confused: if you have a component that say it runs at 12V, it still needs a certain amount of current to run, right? If you add too much resistance, even if you supply a 12V difference across it, the component will not function correctly, right?

That being said, if you don't have the proper voltage (say you only have an 8V difference), but you have a low enough resistance such that the current through the component would be sufficient to power it if the voltage difference were 12V, will the compoennt still run?

I'm asking because I have a windshield wiper pump that's rated at 12V, but isn't running when I power it with an 8V, 1A power supply, even if I don't have any resistance on it (i tried this for less than a second, when 10Ω resistance didn't work). Is this because the current isn't high enough? Or the voltage? Or both?

Any assistance would be greatly appreciated. Thanks!

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  • \$\begingroup\$ possible duplicate of Choosing power supply, how to get the voltage and current ratings? \$\endgroup\$ – Kellenjb Aug 13 '12 at 22:45
  • \$\begingroup\$ The question I am linking to should help clear up a lot of the common misunderstandings around this. If you still have questions after reading through that, feel free to ask. \$\endgroup\$ – Kellenjb Aug 13 '12 at 22:46
  • \$\begingroup\$ @Kellenjb The OP seems to have a fundamental misunderstanding about voltage and current, more so than a lack of knowledge on how to size a power supply. \$\endgroup\$ – Adam Lawrence Aug 14 '12 at 1:45
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I'm new to electronics too but I'm going to try my hand. Essentially, the amount of resistance in the proposed circuit will control both current through and voltage across the pump. When a component says it needs a certain voltage, I.E 12V, it mean's what it says, so the wiper pump was well within it's right's not to work.

Recall that I = V / R. Assuming you directly connected the battery to the pump, V = 12. The resistance of the pump will dynamically change as it functions, for instance, it would change if you suddenly forcibly physically stopped it(you'd likely get extremely high currents if you did that). In the worst possible case (impossible without a superconducting pump), R = 0 hence I = infinity. The current draw increases as the pump requires more power, simply.

Finally, It is difficult to say exactly what you should aim to supply to your pump because ultimately assuming you have the correct voltage, the current will effectively control the 'power' of the pump (P = VI). The 'load' pump will draw the current that it wants to and this can be a problem - It can draw more then the power supply is specified to handle. So you just need a power supply that supplies 12V @ the maximum expected current draw(I.E the pump's maximum current specification). You can go further by adding in a replaceable fuse.

(My first attempt at an answer, feel free to correct me anyone reading - OP, take this with a grain of salt)

Edit: Kellenjb has provided an excellent link which supersedes my pitiful attempt.


Edit2: Regarding your parallel resistors: There's no particular reason to put them in parallel with each other unless you're trying to reach a particular resistance with the parts on hand. As long as they are in series with the pump as a unit though, there is no problem. Here's a few diagrams to clear up any confusion:

Above: Fine, the resistors are all in series, limiting current.

Above: Perfectly fine, the parallel resistors are in series with the pump, limiting current.

Above: Not okay, the resistors are NOT limiting current to the pump]


You asked whether you could damage your pump if it receives too much current - yes you could. But it's important to remember that the pump itself 'asks' for the current. The only function of the resistors is to limit current draw.

Consider if the resistance of the pump, as the sole load in the circuit, dropped to 0, or near to it. You'd get extremely high currents. The point of the resistors is to prevent this from happening or at least reduce the severity when it does. That is also why they are added in series with the pump, otherwise the pump could just draw whatever it wants.

The downside is that you have wasted power. Like I said, check the pump specification, we won't be able to give you an answer on what is and isn't enough current otherwise.

Also to note: I only know this because I'm a computer enthusiast, but pumps can be permanently damaged if you simply provide them power without giving them anything to 'pump' so to speak(so don't 'dry' run it).

We absolutely cannot give you any-more help until we see the pump specifications. Otherwise you really will just be conducting an experiment so to speak.

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  • \$\begingroup\$ Thanks! Both this and Kellenjb's comment made things much clearer for me. I'm not actually using this on my car though. I'm trying to power it from a wall socket. Right now, I'm using a 12V, 1000mA power supply. I addded three 10Ω resistors in parallel, and got the current up to 1.5A (according to my multimeter, anyway), but the pump did not start running. Could it be that this isn't enough current yet? \$\endgroup\$ – Mason Aug 13 '12 at 23:25
  • \$\begingroup\$ On another note, is it possible to damage the pump if I supply too much current to it? Or would the pump have some kind of internal control to make sure that if too much current is inputted, it disperses the extra current and uses only what it needs? (I'm asking if it's a bad idea to test this, very briefly of course, without any added resistance at all, to make sure the pump itself isn't defective). \$\endgroup\$ – Mason Aug 13 '12 at 23:25
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    \$\begingroup\$ @Mason You're getting mixed up. Adding parallel resistors isn't making any current flow into your pump, it's just wasted power. \$\endgroup\$ – Adam Lawrence Aug 14 '12 at 1:47
  • \$\begingroup\$ Sorry, I meant resistors in parallel to one another, all of which are in series with the pump. That wouldn't create a larger current? \$\endgroup\$ – Mason Aug 14 '12 at 3:55
  • \$\begingroup\$ I've tried to clear up some of your remaining questions. \$\endgroup\$ – ColdestShadow Aug 14 '12 at 17:09
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if you have a component that say it runs at 12V, it still needs a certain amount of current to run, right?

The component, when connected to a 12V supply, will draw as much current as it needs to operate. If the power supply being used cannot deliver that amount of current, it may shut down, or the voltage may sag. The voltage sag may prevent the device from running - it all depends on the device.

If you add too much resistance, even if you supply a 12V difference across it, the component will not function correctly, right?

It depends by what you mean by 'adding resistance'.

If you add series resistance, the component will still try to draw current, but the current will generate a voltage drop across the series resistance and reduce the voltage seen by the component. (The total voltage across the resistor and across the component must equal the supply voltage.)

If you add parallel resistance, it will draw current along with your load. This current is wasted as it doesn't help your component whatsoever.

That being said, if you don't have the proper voltage (say you only have an 8V difference), but you have a low enough resistance such that the current through the component would be sufficient to power it if the voltage difference were 12V, will the component still run?

No guarantees. If the component needs 12V to run, don't expect it to work properly at 8V. It may, or may not - it depends on the component's characteristics. There's no general answer.

I'm asking because I have a windshield wiper pump that's rated at 12V, but isn't running when I power it with an 8V, 1A power supply, even if I don't have any resistance on it (i tried this for less than a second, when 10Ω resistance didn't work). Is this because the current isn't high enough? Or the voltage? Or both?

Your major problem is the voltage. If the correct voltage isn't applied, there's no guarantee that the pump will draw any current at all. You're playing with a 12V motor - you really need to start with a 12V supply.

If you apply 12V and the motor drags the supply voltage down to 8V, that's an indication that the supply can't deliver enough current - get a stiffer power supply and try again.

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Let's for the sake of argument assume that the pump behaves like a simple resistor, and that it needs 12 W of power to operate. Since

\$ P = \dfrac{V^2}{R} \$

R will be 12 Ω. If you supply only 8 V then you will get less power:

\$ P = \dfrac{(8 V)^2}{12 \Omega} = 5.3 W \$

which is less than half of what's needed, so the pump won't run. Notice that in the equation voltage is squared, so a small voltage difference may result in a much larger power difference.

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