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I know this is a bad idea in practice, I'm more concerned with the theory. How would you use ohm's law to solve this problem.

Given V = IR

I have a 12v 10amp DC power supply. I want the output voltage to be 5 volts.

So 5 = 10R R = .5 ohms

Is that accurate? I'm very new to electronics but that number seems low to me

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closed as unclear what you're asking by Andy aka, pjc50, Olin Lathrop, Leon Heller, PeterJ Sep 4 '16 at 11:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ If you can arrange things so that \$10A\$ flows through your \$0.5\Omega\$ resistor, then there will be a \$5V\$ drop across it. That's what your computation says in English. But that statement is entirely independent from your earlier statement about having a \$12V\$ @ \$10A\$ power supply, which instead says that you have a \$12V\$ voltage regulated power supply that can comply with up to \$10A\$, if required. The one has nothing to do with the other. \$\endgroup\$ – jonk Sep 3 '16 at 6:32
  • \$\begingroup\$ Possible duplicate of Voltage Divider Question \$\endgroup\$ – pjc50 Sep 3 '16 at 9:52
  • \$\begingroup\$ -1 for the sloppiness with units. You have 10R equated to a dimensionless value, then suddenly it has a value in Ohms in the very next equation. We do engineering here. This sort of nonsense is not tolerated. If you dont care enough to get such basic and simple things right, why should any of us care about your problem either? \$\endgroup\$ – Olin Lathrop Sep 3 '16 at 12:41
  • \$\begingroup\$ Apologies for the sloppiness. I honestly just started my first EE class in my associates and am trying to understand the basics. If this is the wrong place for that I'll gladly ask the question elsewhere. Any suggestions? \$\endgroup\$ – richbai90 Sep 3 '16 at 17:15
  • \$\begingroup\$ It's OK to ask basic questions at the limits of your knowledge. However, it's never OK to be sloppy with units, undefined variables, significant digits, and the like. Those are all basics you were taught in grade school, so there is no excuse. If you don't take them seriously, you will continue to be dismissed as noise and a waste of time. If it's too sloppy to hand in as homework, it doesn't belong here either. \$\endgroup\$ – Olin Lathrop Sep 3 '16 at 21:03
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No, you are wrong.
If you want to change voltage of power source you need some sort of DC-DC convertor (buck convertor in this case).
If you want just a voltage level of 5V, you need just voltage divider (like 7kΩ and 5kΩ).

UPDATE:

So 5 = 10R R = .5 ohms

This calculation is wrong. You MUST use units. "5" of what? "10" of what?
If you mean this:
5V = 10A * R, and you want to calculate R, than it has nothing to do with fact you have 12V power source. Formula "5V = 10A * R" means that you apply 5V at unknown resistor, will measure 10A current and want to calculate that unknown resistor. But if you connect 0.5Ω resistor to 12V source, you will need to provide 24A of current.

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I have a 12v 10 amp DC power supply. I want the output voltage to be 5 volts. So 5 = 10R R = 0.5 ohms.

Your idea will work only if the load is constant such as a resistor, a bulb or a heater. If the load changes the 5 V will change too. 5 V implies you are running a micro controller or some kind of logic. These are not steady loads, are quite sensitive to voltage changes and you could easily destroy your device.

When a constant voltage is required a voltage regulator is used. This will adjust it's output to keep the voltage at the required value.

enter image description here

If you only need a small current (100 mA, for example) a 7805 voltage regulator will do the trick. (Don't skip the decoupling capacitors.) If you require more current you should consider a switching regulator as these are more efficient and won't dissipate at much heat.

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If your '10A' supply switches into a constant-current mode like a typical lab bench supply, and it is set to exactly 10A for the current limit, then it will give you 5V out with a 0.5 ohm resistor. When you first connect it, it might give you a big spike because of internal capacitance.

It will also get extremely hot (both the resistor and the power supply) and is a 5V supply with exactly 0% efficiency. You could load it with say 500mA and the voltage would only drop 5%, at which point you might have 80W being consumed to produce an output of 2.5W so our efficiency improves to a whopping 3%.

However most supplies are not like that, and will limit at considerably more than their rating for period of time, so you might see it throbbing at 8V or something of that ilk. Thus promptly destroying anything that was limited to 5 or 6V and connected to the output, of course.

The rating of a power supply is what you are allowed to draw long term and only loosely related to what actually happens when you apply a load.

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I have a 12v 10amp DC power supply. I want the output voltage to be 5 volts.

Then get a 5 V supply. Or, you could follow the 12 V supply with a buck regulator.

So 5 = 10R R = .5 ohms

This is obviously wrong just from looking at the units. R is a dimensionless quantity in the first equation, then suddenly in units of Ohms in the second.

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You are correct assuming that you have some load that will draw precisely 10A from a 5V supply. Then the supply will produce 10A, the resistor will drop 5V, and that will leave 5V across the load.

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  • \$\begingroup\$ It's a 12V supply. If the load is also 0.5 ohms,both resistors are identical, you'll have 6V across each. Unless the supply is current limited to precisely 10A... \$\endgroup\$ – Brian Drummond Sep 3 '16 at 10:42

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