It's been a few years since I first studied the Ohm's law at school. Now I'm really getting into electronics, and I must admit a part of it still puzzles me.

I know this is kind of a strange question, but I have no clue how to express it otherwise, so I'll just lay out my reasoning. The following train of thoughts is wrong and/or incomplete. Please stop me whenever I'm wrong (even for the sligthest thing), and complete this reasoning.

Here it comes:

Let's make a circuit with:

  • 6V battery (typically 4 AA batteries)
  • 2k ohms resistor
  • LED, for the sake of doing something

Circuit map:


simulate this circuit – Schematic created using CircuitLab

From Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

\$ I = \dfrac{V}{R} = 3mA \$

I do understand that the current is limited due to the resistor... crystal clear.

The book tells me that, if I measure the potential difference between both sides of the resistor, I will have something between 0 and 6 volts.

Sounds legit:

  • can't be 6 volts because the two ends of the battery are connected, therefore lowering the potential
  • can't be 0 volts, since the current does not freely move around, due to the resistor; resulting in one side being more negatively charged, and the other more positively charged, since the electrons can move faster to an end of the battery than through the resistor.

What I don't understand (and if my whole reasoning is right until now), is: How can I calculate the expected voltage around the resistor ?

  • \$\begingroup\$ For the next time: we have an on-site schematic editor, you can launch it with Ctrl-M. \$\endgroup\$
    – user17592
    Apr 16, 2013 at 16:52
  • \$\begingroup\$ Brilliant ! Will do :) \$\endgroup\$ Apr 16, 2013 at 17:01
  • 1
    \$\begingroup\$ I'd like to clarify that the question is not an exact duplicate of the linked one. Just, it provides a complete and thorough explanation about how circuit with LEDs behave, since your question - while being about the resistor - is strongly dependent on the LED behavior. I hope you can find there the answer to your question, otherwise you can tell what's missing and ask for reopening. \$\endgroup\$
    – clabacchio
    Apr 16, 2013 at 20:15
  • \$\begingroup\$ Excellent, thanks for the link and for adding the schema! \$\endgroup\$ Apr 17, 2013 at 8:26

3 Answers 3


From the Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

Na-ah, the LED has a resistance too. Common LEDs use 2.5V at 20mA. We can calculate the LED's resistance (on 2.5V) with \$R=\frac{V}{I}=\frac{2.5}{0.02}=125\Omega\$.

Now, when it comes to your circuit, you shouldn't just use a 2K resistor, you should think about that and calculate the needed value. Now, the voltage over the LED probably won't be 2.5V, which means the LED has a different resistance, different current is flowing, etc. You cannot calculate the current flowing in this circuit and therefore you cannot calculate the voltage over the resistor (we'll come to that later). Here's how you should design your circuit:

You want a 0.02A current through the circuit (the current is the same everywhere as it is a series circuit). That means the total resistance has to be \$R_{tot}=\frac{V}{I}=\frac{6}{0.02}=300\Omega\$. We already saw the resistance of the LED was \$125\Omega\$, that means your resistor has to be \$300-125=175\Omega\$.

Now, how do you calculate the voltage over the resistor? Take Ohm's law, the current through the resistor (0.02A) and the resistance (\$175\Omega\$):

$$V=I\cdot{}R = 0.02\cdot175 = 3.5V$$

Note that this also equals \$6V-2.5V\$.

  • \$\begingroup\$ Excellent answer! Many thanks. For the sake of curiosity, what happens if the resistor was less than 175 ohms? Would that create a so-called "short circuit", because not all voltage is taken? On the opposite, what happens to the voltage if I insist in using this 2k resistor? It really cannot be calculated? \$\endgroup\$ Apr 16, 2013 at 17:20
  • \$\begingroup\$ @Antoine_935 "not all voltage is taken" - no, all voltage will be taken, but because of the lower resistance, the current will increase (V=IR). So it might damage your components. Your LED will be brighter though. The problem with the 2K is that it falls out of the LEDs specs. LEDs don't follow Ohm's law, so the specs can only be applied on the intentional circumstances. It really cannot be calculated thus, but you can of course just measure it with a voltmeter. \$\endgroup\$
    – user17592
    Apr 16, 2013 at 17:22
  • \$\begingroup\$ Waw, some things do not follow ohm's law? Totally ignored that. I guess I'll ask another question "only" for this. Again, many thx! \$\endgroup\$ Apr 16, 2013 at 17:25
  • \$\begingroup\$ @Antoine_935 it's indeed best to ask a separate question for that. And you're welcome, of course :) \$\endgroup\$
    – user17592
    Apr 16, 2013 at 17:26
  • 11
    \$\begingroup\$ It is bad practice to calculate the effective resistance of an LED as it varies wildly with current. It is better to model an LED with a voltage drop. \$\endgroup\$
    – jippie
    Apr 16, 2013 at 18:13

The sum of all voltage drops must equal the source voltage.

\$V_{source} = V_{resistor} + V_{led}\$

Depending on what wavelength the LED the operates at, it will essentially drop the same the voltage (when you are within the current restrictions of the LED).

If we assume that the LED drops 1.2V, then that means that the resistor must drop 4.8V.

The LED forward voltage will always be a constant (might deviate slightly), when you are within the current limits.

If the LED can take a max of 20mA, then at 15mA and 10mA, the voltage drop across the LED will always be the forward voltage drop (in this example, the 1.2V).

Make sense ?

  • \$\begingroup\$ Nice answer, it's helping. I'd upvote it too if I could, thanks. \$\endgroup\$ Apr 16, 2013 at 17:21
  • \$\begingroup\$ The forward voltage is not constant. As LEDs heat up, the forward voltage drops and the current passing through the LED increases, which creates additional heat. This can lead to thermal runaway and even destroy the diode. This is why constant current source are preferred when driving LEDs. \$\endgroup\$
    – Rev
    Apr 16, 2013 at 20:41

Your LED will have about 2 volts across it for a standard LED which means you'll have about 2mA flowing in the circuit. It won't be very bright.

When you said: -

I = V/R -> 3mA

That's where you made an error. Even though the LED may be talked-about as having little resistance it still drops a voltage across it and for your average LED it will be about 2V whether you're feeding it with 2mA or 20mA.

  • \$\begingroup\$ Nice answer, it's helping. I'd upvote if I could, thanks. \$\endgroup\$ Apr 16, 2013 at 17:20
  • \$\begingroup\$ Note that the actual forward voltage depends on the forward current. This is why most diode data-sheets provide a "fwd voltage VS current" graph. From the answer is sounds like this isn't the case. \$\endgroup\$
    – Rev
    Apr 16, 2013 at 20:24
  • \$\begingroup\$ @Rev1.0 - it is simplistic i agree \$\endgroup\$
    – Andy aka
    Apr 16, 2013 at 20:54

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