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I've come across two ways to determine which resistor value is needed for a load. The first one, which is always titled 'How to determine resistor value for an LED' says that you can use the formula:

$$V_{in} = input\ voltage$$ $$V_{l} = LED\ voltage$$ $$I_{l} = LED\ current$$ $$R = \frac{V_{in} - V_{l}}{I_{l}}$$

And I've also come across one that says to use resistors in series to add a voltage drop like this:

schematic

simulate this circuit – Schematic created using CircuitLab

And then use the formula:

$$V_{out} = V_{in} \cdot \frac{R2}{R1 + R1}$$

My final goal is an attempt to power the Raspberry Pi (5v 2.1A) with a 12v battery pack. So, could I just use a 3.3R resistor:

$$R = \frac{12 - 5}{2.1}$$ $$R = 3.333$$

Or do I need to use the second equation to drop the voltage? I'm just confused as to which method to use

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  • \$\begingroup\$ The first formula works only for LEDs. The second formula works only for high-impedance loads. The Pi is neither. \$\endgroup\$ – CL. Nov 1 '15 at 12:16
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You will instantly destroy the Raspberry Pi with a 12V battery and series resistor. The Raspberry Pi requires 5V regulated at as much as 2A (but usually much less). So the series resistor will not drop the expected voltage, and the Pi will be dead.

You should use a switching regulator to drop the voltage (actually to output a regulated 5V from the 12V-ish input)- you can buy inexpensive modules that have the chip, inductor and other parts, all for a few dollars.

http://www.compoexpress.com/media/catalog/product/cache/2/image/500x500/e6f6636d1b4b3c56f73fd0ae52afb1ff/l/_/l_r5g6lm2596s-dc-dc-adjustable-step-down-power-supply-module_1_.jpg

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    \$\begingroup\$ But theoretically, if the 12v source was stable, and it was a perfect 3.3R resistor, would this be the correct way to set up the circuit? \$\endgroup\$ – Just Lucky Really Oct 31 '15 at 17:50
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    \$\begingroup\$ Your calculation is correct, but you would also need to guarantee that the load was a perfectly steady (within a few percent) 2.1A. If all those conditions were true then it would be okay. In order of increasing problems the sources of trouble are the resistor (best), the '12V' and then the variable and unknown load (worst). \$\endgroup\$ – Spehro Pefhany Oct 31 '15 at 17:53
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    \$\begingroup\$ @Stretch - even if you could use a drop resistor (which you can't for the reasons shown above) you would also need to consider how much power the resistor would dissipate. With a 7V drop at 2 amps, the resistor would be dissipating 14 watts. That's power you're wasting and heat you're creating. \$\endgroup\$ – DoxyLover Oct 31 '15 at 18:05
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    \$\begingroup\$ Yeah, I didn't take that into consideration when I gave it a go earlier (Before posting this question) ... I tried using a 3.3R resistor that was only rated at 0.6W ... I now have a resistor shaped burn on my thumb \$\endgroup\$ – Just Lucky Really Oct 31 '15 at 19:17

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