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As the title says, do I need trivial resistor or not if V/I = zero or almost zero?

For instance:

Assume I'm using 3V buzzer and 2V LED in series with an Arduino 5V pin.

$$R = \frac{(3 + 2 - 5)V}{15mA} = 0 \Omega$$

Should I add a trivial resistance just in case, or not, and why?

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    \$\begingroup\$ As clever as that sounds, you'll see the problem with it if you turn it around. \$\endgroup\$ – Ignacio Vazquez-Abrams May 6 '14 at 18:14
  • \$\begingroup\$ What kind of buzzer are you using? Piezo or coil-based? Does the buzzer have a driver circuit already? With your comment that all 5V is dropping over the Buzzer, that could imply the buzzer is either broken, you don't have a connection to the LED, or you don't have a driver circuit built into the buzzer. \$\endgroup\$ – horta May 6 '14 at 18:59
  • \$\begingroup\$ related (duplicate pretty much): electronics.stackexchange.com/questions/12865/… \$\endgroup\$ – Nick Alexeev May 6 '14 at 22:27
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You can try to do this with the buzzer, but you have no idea what current the buzzer wants. If it wants 15 mA, have a field day and give this a go. You won't need the resistor. You can try it, anyway. Even if the numbers aren't perfect, you'll either hit an equilibrium point where the buzzer load line intersects with the LED curve, or you won't forward bias the LED enough to make it turn on, and no current will flow.

I recommend wiring the buzzer and resistor in parallel, assuming you won't pull too much current from the Arduino pin (checking this is on you!). If you're pulling too much current, use a transistor.

If you want 15 mA going through a 2V LED, you need to drop 3 volts.

$$R= \frac{3V}{15mA} = 200 \Omega$$

You'd use the same calculation for the 2 volt drop you want for the buzzer, after you figure out how much current it needs.

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    \$\begingroup\$ So then the question becomes "Why isn't the 3V buzzer dropping 3V?". \$\endgroup\$ – Ignacio Vazquez-Abrams May 6 '14 at 18:25
  • \$\begingroup\$ Got it, @IgnacioVazquez-Abrams, and was correcting as your comment came through. I missed the point, but the answer is correct now. Thanks! \$\endgroup\$ – Scott Seidman May 6 '14 at 18:28
  • \$\begingroup\$ their should be 2V drop on led terminals and 3V drop on buzzer terminals with 15ma going in both which is exactly what i need, so why would it be wrong. edit : sorry didn't read your edit, thanks buddy \$\endgroup\$ – Andrew May 6 '14 at 18:29
  • \$\begingroup\$ replying to the new edit about in parallel, the micro/arduino pins only get 25 mA which may not be enough for both, that's why i thought of it as series in the first place, \$\endgroup\$ – Andrew May 6 '14 at 18:32
  • \$\begingroup\$ @AndrewxXx -- I edited yet again! Give it a try. Can't hurt anything, but don't be surprised if it doesn't work. \$\endgroup\$ – Scott Seidman May 6 '14 at 18:35

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