I have been very confused with resistors because I have not found a great, although simple explanation about if resistors reduce or limit voltage. Also, I want a explinaton on why you subtract Vf from V when finding the right resistor for a LED. I get that V=IR, but I don't get how to get the resistor value. I take (V-Vf)=I*R, that means that if I want to get 5v to 3.3v@25mA, I take 1.7=0.025*r, meaning that the resistor would REDUCE VOLTAGE by 1.7 volts, correct? Thanks! *Feel free to make any comments and questions, this is quite vague.


Resistors resist the amount of current flowing through them. When they do this, a voltage difference is developed across the resistor.

LED are usually characterized by the amount of voltage & current they need in order to light up. So a 3.3 volt 25mA LED needs to develop 3.3 volts across the LED.

In closed loop circuit all the voltages need to add up. This is Thévenin's theorem. So when calculating which resistor is needed for a given constant voltage power supply, say 5 volts, and given LED which requires, say, 3.3 volts, you subtract 3.3 from 5 and get 1.7. See, the power supply adds 5 and the LED takes away 3.3 and then the resister needs to take away the remaining 1.7. All the while you want 25mA running through the loop. So, using your equation you have:

5 - 3.3 = I x R
1.7 = 0.025 x R
R = 1.7 / 0.025
R = 68 ohms

Do check to make sure these calculations are correct and that you really want 25mA running through the LED. Most LEDs these days take only a fraction of that amount of current!

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  • \$\begingroup\$ Thank you SO MUCH! Extremely helpful, easy to understand and informative. I really hope this helps others, too! \$\endgroup\$ – Blake Jan 6 '17 at 0:45
  • \$\begingroup\$ One question: Let's say I plugged in 2.7V instead of 1.7V into V=IR, Would that leave me 1.0V left after the 3.3 voltage drop of the LED if I used the returned resistor value? \$\endgroup\$ – Blake Jan 6 '17 at 0:52
  • \$\begingroup\$ The voltage and current in a circuit is a balancing act. You can not have "left overs". When you go around a complete loop of a circuit, the voltage needs to add up to zero. And the same current needs to be flowing through every part of that loop. If you raise the resistance of the resistor you might get the voltage across the resistor to increase from 1.7 to 2.7 volts. But all the while less and less current will flow through the loop. And the LED will get dimmer and dimmer. \$\endgroup\$ – st2000 Jan 6 '17 at 3:04
  • \$\begingroup\$ Thanks, couple things: does 'voltage across the resistor' mean the voltage lost while going through the resistor? Also, if I only want to pull 3.3v, does it only pull 3.3v? You said there couldn't be any 'leftover', although in my Raspberry Pi, if I pull 5v and then hook up a 3.3v LED to it, I thought 1.7v would be 'leftover' and go to ground. Sorry for being a bit vague ;) \$\endgroup\$ – Blake Jan 6 '17 at 6:28
  • \$\begingroup\$ It is more like the resistor is a constriction in the water plumbing and the pressure before the constriction is 3.3 lbs/sq-in more than after. For more about using plumbing as an electronics metaphor try reading this. Or looking at my answer for this stack-exchange question. \$\endgroup\$ – st2000 Jan 6 '17 at 19:48

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