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That's something I've seen in many places while talking about power supplies, but something I've never understood. At a very basic level, the current flowing through a load is inversely proportional to the load resistance. So how does a device rated at 50 mA only draw 50 mA from a 5V/1 amp power supply, and how does load resistance factor into it?

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    \$\begingroup\$ "So how does a device rated at 50 mA only draw 50 mA from a 5V/1 amp power supply, and how does load resistance factor into it?" What kind of answer are you looking for here when you ask "how"? Because normally this is answered by V=IR but I don't think that is what you are after. Keep in mind that most things are not resistors. \$\endgroup\$
    – DKNguyen
    Oct 24 at 19:56
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    \$\begingroup\$ No, most things are not pure resistors. There is always resistance, but there are other aspects in the device that affect current draw as well. Such as diode behaviour, and back EMF. \$\endgroup\$
    – DKNguyen
    Oct 24 at 20:00
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    \$\begingroup\$ > will using a 10V power supply draw double the current? -- probably not. It could, if it is a purely resistive device. But probably it is not. The current could go up, but not linearly (incandescent lamp), it could go down (switchmode LED power supply), it could stay the same (linear LED supply). It depends on what the device is. \$\endgroup\$ Oct 24 at 20:06
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    \$\begingroup\$ @Soumil There is no possible answer in electronics. Electronics has a very high level view for the most part. The answers as to why will be found only in a physics understanding. When power is applied, charges very rapidly move along the surfaces of conductors and other materials (including insulators, which are only slightly affected) and self-arrange for reasons that can only be followed by understanding physics. This self-arrangement then impels currents. Electronics does not study this effect, nor does it include it in its body of work. That study is outside the field. \$\endgroup\$
    – jonk
    Oct 24 at 20:11
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    \$\begingroup\$ Confused on which duplicate this has to be tagged to as there are tons. \$\endgroup\$
    – Mitu Raj
    Oct 24 at 20:26
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At a very basic level, the current flowing through a load is inversely proportional to the load resistance.

Not really. For some loads, where current and voltage are generally proportional to each other, we define its resistance as the ratio of voltage to current. Pure metal wires are like this. Other loads don't have this proportionality, like diodes, or plasma arcs.

Think of a 5 V 1 A power supply as a water tank 5 m above ground (the voltage), with a large enough pipe coming from it to be able to deliver 1 litre per second (the current capability).

If we connect a very thin pipe to it, then that pipe might only take 1 ml per second with 5 m head of water across it. A thicker pipe could draw 10 ml per second. The load pipe is drawing only the current it needs, not the maximum flow delivery possible from the tank.

Another way to look at this is to say that as you try to force more water through the pipe, the pressure across it increases. Once you are forcing 1 ml per second through it, the pressure difference has risen to 5 m head of water, where it equals the pressure available from the tank.

The flow through a pipe tends to be proportional to the pressure, so we would define a 'resistance' to it. Flow of water across a leaky weir is more like that of a current through a diode, little until some voltage is reached, and then increasing rapidly with every further small increase in voltage. You have to force an awful lot of water across a weir to get the pressure across it to rise significantly.

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the current flowing through a load is inversely proportional to the load resistance

That's only true of resistive loads (such as a room heater). It is untrue of most loads (such as a power supply, motors, LED, etc.). Most loads do not have a fixed value of resistance, so the current is not directly proportional to the voltage.

enter image description here

So how does a device rated at 50 mA only draw 50 mA from a 5V/1 amp power supply

Because that's what it needs. The power supply can provide any current from 0 to 1 A, whatever the load wants. It does not force 1 A out. If the load wants 1 A, fine. But if the load wants only 50 mA, that's fine too.

As an analogy, if an ATM can only give you up to $ 400, it doesn't mean that it must spit out $ 400. If you only want $ 20, you just take $ 20. It's up to you.

Same with a load: it takes what it wants.

how does load resistance factor into it?

It probably doesn't at all: a device rated for 50 mA is likely not a resistor. Hence, it doesn't have a "load resistance".

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    \$\begingroup\$ "Because that's what it needs. The power supply can provide any current from 0 to 1 A, whatever the load wants. It does not force 1 A out. If the load wants 1 A, fine. But if the load wants only 50 mA, that's fine too." I guess I'm more after what's going on internally here for the load to "want" only 50mA. \$\endgroup\$
    – Soumil
    Oct 24 at 20:02
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    \$\begingroup\$ > what's going on internally here for the load to "want" only 50mA. -- Then please edit your text and ask that question. \$\endgroup\$ Oct 24 at 20:03
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    \$\begingroup\$ It doesn't "want" 50mA as much as 50mA is what it requires to do its job when we designed the circuit. That circuit require X Volts to push 50mA through it. If you apply more voltage, more current (sometimes excessive) does flow. This is like asking "Why does a bicycle want you to pedal at some RPM and force to move it at some load and speed?" \$\endgroup\$
    – DKNguyen
    Oct 24 at 20:04
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    \$\begingroup\$ @Soumil For a very basic understanding of the concept, you can think of an electronic device adjusting its own resistance, or a lightbulb's resistance changing when it heats up. (And then there are things that just aren't resistors to begin with, like capacitive loads) \$\endgroup\$
    – user253751
    Oct 25 at 11:04
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  • Think of voltage as water pressure.
  • Think of current as water flow
  • Think of resistance as the diameter of a pipe combined with how much you turn the tap
  • Think of filling your bath/tub

How does you bath not fill instantly you turn on the tap (infinite current) despite the fact the reservoir (PSU) could also fill all the baths in the street at the same time?

Why does your bath fill at a constant rate (if the pressure remains the same)?

Answer: That fill rate (current) is determined by the pressure (PSU voltage) and pipework (resistance) and is nothing to do with the capacity of the reservoir (PSU current rating)

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  • \$\begingroup\$ How is this answer any different from the already posted water pipe analogy? \$\endgroup\$
    – Lundin
    Oct 25 at 10:57
  • \$\begingroup\$ It's far shorter and to the point, feel free to upvote which ever you choose. \$\endgroup\$
    – Jason M
    Oct 25 at 12:30

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