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schematic

simulate this circuit – Schematic created using CircuitLab

Let's say I have a power supply that supplies 5v and 3A.

If I add a 2 ohm resistor, it could do 2 different things.

Realizing that resistance = volts/amps...

either it will drop the voltage by 1 volt and give 0.5A, as 1/0.5 = 2 ohms

OR it will drop the voltage by 4 volts and give 2A, as 4/2 = 2 ohms.

Let's just assume the power rating on the resistor is high enough that we don't have to worry about it for this example.

Question

So which one of these two scenarios will become the reality if I have just the power source and a single 2 ohm resistor? And why? Otherwise, if you suspect there is some concept I fail to misunderstand, please explain that as well. I'm basically saying, according to ohm's law it seems both these scenarios are possible when deciding a voltage drop and current to find a resistor value. I'm trying to ask a basic question to understand resistors, not asking because of a real life application.

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    \$\begingroup\$ Do you mean there is a power supply designed to provide \$5\:\text{V}\$ and that it is already loaded with a fixed resistance that results in \$3\:\text{A}\$ and that you will now insert a \$2\:\Omega\$ resistor in series with the existing load? Or in parallel with the existing load? Or something else? What do you mean by "add," I guess. What is the existing "circuit" you are talking about? \$\endgroup\$ – jonk Mar 29 at 5:19
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    \$\begingroup\$ You fail to understand that your terms are ambiguous, and that we can't see the picture in your head. Add a schematic for the circuit before you add the 2ohm resistor, and after adding it. Hit the edit button below your post, then the resistor/pencil/diode/capacitor button on the edit box toolbar to bring up a schematic editor. \$\endgroup\$ – Neil_UK Mar 29 at 5:43
  • \$\begingroup\$ Your power supply is a constant voltage one, meaning you have two knowns, U and R, and one unknown, I. \$\endgroup\$ – winny Mar 29 at 8:39
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If the supply potential is 5V and can delivery up to 3A then a 2Ω load will draw only 2.5A (=5V/2Ω) and still see 5V.

This may continue if the load is reduced to 1.66Ω (=5V/3A) but any lower may shut down the supply.

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  • \$\begingroup\$ Thanks, I realized the voltage drop across that one resistor would be 5V, leaving it at 0 at the end, a Kirchhoff law. Then I realized that I had to look at voltage drop across more than one resistor in series to learn what I was missing. \$\endgroup\$ – agileOne Mar 29 at 7:52

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