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Why does voltage even matter, if it's the current that goes through a circuit? I understand the relationship V = IR (I think) and that increasing the voltage (pressure) will increase the current.

However, there are some device there that requires a certain amount of voltage, say 5v. My question is why can't you connect it to a 10v power supply and add a 2-ohm- resistor so the current will be the same as the current a 5v will produce? Via I = V/R

I know this probably sounds very stupid, I know it's not going to work that way, and you can still potentially destroy circuits, if you only care about current and not voltage, but I'm just wondering where is the gap in this logic?

I tried it on a circuit simulator website, and the device blew up everytime from the bigger voltage source even though the currentwas the same with the lower voltage source (by adding resistors). (The lower voltage source works just fine)

Unless of course it's not just the current that matters, and that voltage matters more than just producing a certain amount of current

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    \$\begingroup\$ How do you know the resistance of the device you're trying to power? \$\endgroup\$ – immibis Jul 12 '18 at 6:12
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Your approach (10V and 2 Ohm series resistor) works (only) for a circuit that draws a constant current of 2.5A at 5V.

When the circuit draws more or less current it will see a voltage different from 5V. For instance, the circuit could behave like an 8Ohm resistor. The total resistance would be 10Ohm (they are in series), so the current would be 1A. The circuit would get 8V (1A * 8 Ohm). That will probably cause it to malfunction, or even damage it.

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My question is why can't you connect it to a 10v power supply and add a 2-ohm- resistor so the current will be the same as the current a 5v will produce? Via I = V/R.

You can do this if the load remains constant. A heater or light bulb, for example, can work like this but even in these simple cases the resistance will be low - by a factor of 5 to 10 for the light bulb - when cold so the voltage division will not work out to be exactly 50% until the heater / bulb warms up. For other devices such as motors the current will vary with the load and for electronics the current can vary with the task in hand varying from miniscule while asleep or idle to relatively high while running.

Your 2 Ω resistor example will cause a voltage drop of 2 V per amp and would drop 5 V only when the load current is exactly 2.5 A. At lower current the voltage on the device would rise. At higher current the voltage on the device would fall. The voltage regulation would, most likely, be unacceptable.

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There's a very common misunderstanding about \$V=IR\$. Specifically, what voltage the V is referring to in the equation. In Ohm's Law, the V is precisely the voltage drop across the resistor R that has current I going through it.

The common misinterpretation is that the V refers to the total voltage of the power supply being used. In your example, that is not correct.

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Voltage matters, if you're running circuitry at a higher potential then what it is designed for then you can expect components to overheat and sustain damage as they are run beyond there maximum ratings and conditions.

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Unless of course it's not just the current that matters, and that voltage matters more than just producing a certain amount of current

In real circuits both matter. An IC, for instance, needs a certain minimum voltage to work (silicon diodes, for example, don't conduct at all below about 0.7V). On the other hand, too high a voltage may destroy semiconductor structures just by the force it exerts on the electrons. (Really high voltages can even cause arcs through normally insulating parts, permanently damaging the insulation. Much lower voltages can cause similiar effects through the miniscule insulating structures in ICs.)

Then, as per V=RxI, or I=V/R, higher voltage often forces more current through a given conducting structure, which technically is mostly not a problem by itself, but a higher current through any non-superconductor causes higher power loss in the conductor which heats the conductor and may ultimately cause irreversible thermal damage.

So, practically, we need to maintain the voltage between the lower and upper bound for the circuit to operate. Too low and it won't work, too high and it may burn out.

As others have answered, most non-trivial circuitry does not act like a constant (ohmic) resistance but varies its apparent impedance over time. Digital ICs (CMOS), for example, often briefly consume much more power on the edges of their clock signal than in between, so powering via some sort of constant current source will force higher voltage and too much current through them at times and/or not enough at other times.

The analogy with water (pressure=voltage,flow=current) holds: If you had a source that always forces as constant flow (current) of, say, 1 liter per second (Ampère) through your piping, what would happen if you shut a valve between the source and the drain? The source would increase the pressure until the flow is at 1 liter per second again, which may happen only after the (shut) valve or other parts of the piping break. Also, as the flow through all of the piping combined is fixed at 1l/s, flow and pressure in different parts of the piping will fluctuate in response to other (parallel) parts of the piping changing the flow through them, which is undesirable in electric circuits.

As to the question in the heading,

Why can't you just reduce the current instead of the voltage?

That's basically what we do all the time. Power supplies keep the voltage constant and modulate the current in response to the current the powered circuit draws (its resistance) over time to always keep the output voltage constant. The voltage stays at e.g. 5V even when you disconnect the circuit (=0A).

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