In this video, https://www.youtube.com/watch?v=ks_VShSgPns, a 220 ohm resistor is connected across a 5V DC source (the source happens to be oscillating between 0V and 5V but that's besides the point).

I would think that, by ohm's law, the current through the resistor is I = V/R = 5/220 = 22.7mA, and I'm sure this would indeed be the case with an ideal voltage source (I think he demonstrates it at some point in the video).

However the point in this video is to illustrate that the voltage provided across the black and yellow wires cannot push enough current and is a "bad source", and so when measuring the actual current at 0:58 in the video, the reading is only 0.29mA.

This reading clearly does not obey ohms law, as the resistor would technically have a value of around 17k ohms in this case.

My question is, what is exactly happening here? Is the resistor simply not being "put to test" because the source cannot push current in the first place? Or why is ohm's law not being obeyed?

  • 1
    \$\begingroup\$ Hmm... ohms law is wrong.... or maybe something else? \$\endgroup\$
    – Andy aka
    Commented Sep 1, 2020 at 18:02
  • 3
    \$\begingroup\$ Sorry, but that's a > 8 minute video so not many of us are going to watch it for you. Instead, you should post a screengrab of the relevant section. That way the question is also understandable if the video is removed from YouTube. \$\endgroup\$
    – Transistor
    Commented Sep 1, 2020 at 18:04
  • 3
    \$\begingroup\$ I did not watch the video but usually the answer to this question is the following: The bad source is probably a batteries have an output impedance that is too high. SO the more current you try to draw from it, the more voltage is dropped internally in the battery before making it to the output. In other words, there's a second hidden series resistor in the circuit. \$\endgroup\$
    – DKNguyen
    Commented Sep 1, 2020 at 18:05
  • 3
    \$\begingroup\$ Ohm's law relates the resistor's current to the voltage across the resistor. Is that voltage shown? Does the author of the video claim it is 5 V? \$\endgroup\$
    – nanofarad
    Commented Sep 1, 2020 at 18:05
  • \$\begingroup\$ I didn't think it was necessary to include specifics on the video, as I expanded it myself. Also, the voltage is shown, yes. Measured as just above 5V \$\endgroup\$
    – Gary Allen
    Commented Sep 1, 2020 at 18:32

4 Answers 4


The internal resistance of the source is responsible for the low current.

Source voltage = 5.15 V

Current limiting resistor value = 220 Ω

Current = 0.00029 A

Total resistance in circuit = 5.15/0.00029 = 17759 Ω

Internal resistance of the source = 17759 - 220 = 17.5 kΩ

  • \$\begingroup\$ That actually makes complete sense, thanks! \$\endgroup\$
    – Gary Allen
    Commented Sep 2, 2020 at 6:27
  • \$\begingroup\$ Anytime, Gary! Many thanks. \$\endgroup\$
    – vu2nan
    Commented Sep 2, 2020 at 13:26

I just looked at the first minute or so of the video.

It seems that the video is a continuation of another video where it was found that the source of the pulsing signal could not provide enough current to light an LED brightly. I assume that the source is a microcontroller output or some other current limited source, so the actual source looks like a 5 volt supply with a 15K resistor in series.

In that video, he is showing the use of a transistor to drive an LED from a low-current signal.


When you measure the voltage at the resistor with no load (i.e., with zero current) it's a bit higher than 5V. But he's measuring the current through the resistor into essentially a short circuit -- and he's not measuring the voltage at the same time.

So, the source (presumably a microcontroller or other logic pin) is really wimpy, and cannot supply more than a few hundred microamperes. If he were to simultaneously measure current and voltage (or just short the resistor to ground and measure voltage) I think you'd find that the resistor voltage was only a few hundred millivolts in that case.


Do not worry about Ohm's law.

What yo do nor see is that the Author a video is using a "signal gen" (NE555 timer) + series resistor to make a "bad source" to model "real digital circuit".


simulate this circuit – Schematic created using CircuitLab

And he disconnects the LED and connects a voltmeter to measure the Open circuit voltage. Then connects the ammeter to measure the Short circuit current.


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