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My question is originally from an youtube video, which is a tutorial series on digital electronics.

In a nutshell, He (the one in the video) has a predesigned input signal which fluctuates the voltage across the terminals, connected to a resistor. Now, he measures the voltage across it and finds that the voltage rises and drops from $5$ to $0$ volts (he measured it in the previous video of the mentioned link).

It is very clear that the resistance of the resistor is $220$ ohm. So the peak current across the resistor according to ohms law must be $23$mA. but look below the images on what he measures,

5V

0.29mA

He measures a current fluctuation of $0.29$mA to $0.00$mA

He have another circuit below which is more like a battery connected to resistor and indeed finds $23$mA as the peak current.

I think there is something seriously wrong with my understanding of circuits, why must the current across the resistor be different(apart from the one that is obtained through ohms law), provided that there is a corresponding voltage between it?

One way many people think about it is in the follwing way, think of current as amount of flow of water, and voltage can be thought of as, how strong is the water being pushed, so from this analogy, if water is pushed across a narrow tube, no matter what, water will pop out the other side the same amount as it is fed into it.

Applying it, can we conclude that the input signal is not supplying enough current across the resistor?? if that is the case, then how do you correctly apply ohms law for such a resistor?

**It would be helpful if you spot where i am missing in understanding ohms law..searched a lot in internet, finally ended up here.

EDIT: I think the most problematic part in the top circuit is the input signal, i can't really think of any voltage source that makes a resistor apparently disobey ohms law, would be helpful if someone spot such one here.

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  • $\begingroup$ Watching the video, it seems that he doesn't measure the resistance on that resistor, so it may be actually lower than 220 ohm. Furthermore, there are uncertainties on the measurements to take into account; I think he should've set the amperimeter to measure in $\mu A$ instead, but I'm no experimentalist. $\endgroup$
    – Renan Nobuyuki Hirayama
    Commented May 16, 2020 at 17:44
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    $\begingroup$ sorry for that, this circuit is shown in two videos, in one of the videos, he actually measures the resistance and finds that it it 220ohm. (i said that this was from a tutorial series) $\endgroup$
    – user243016
    Commented May 16, 2020 at 17:46
  • $\begingroup$ being an experimentalist or not doesn't matter i think so, are we understanding ohms law or not does matter...somewhere in the internet i saw that a 12V car battery supplies more current than a usual 12V battery, i really doesn't understand what that "supply" word actually means.....i am hopeless with the internet on this! $\endgroup$
    – user243016
    Commented May 16, 2020 at 17:51
  • $\begingroup$ I can think of two things it might mean: (1) the voltage is actually higher than $12V$ or (2) the internal resistance is smaller than other batteries, I would guess the latter $\endgroup$
    – Renan Nobuyuki Hirayama
    Commented May 16, 2020 at 17:54
  • $\begingroup$ Also, Ohm's Law is only an approximation, as the answers on this Quora post state: quora.com/When-is-ohms-law-not-applicable $\endgroup$
    – Renan Nobuyuki Hirayama
    Commented May 16, 2020 at 17:56

3 Answers 3

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The 5V is the voltage of the input signal without any load. The source that provides the input signal is not a perfect power source but has an internal resistance.

Later in the video, he actually measures the voltage with the LED and resistor in place and gets a voltage of 1.8V. This means that, as current flows through the circuit, there is a voltage drop in the circuit supplying the input signal.

Ohm's law is still valid but the voltages in the closed circuit are not the same as the voltages in the open circuit.

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The voltage measured in the top photo is not the voltage across the resistor, it's the open-circuit signal voltage (the voltage across the signal leads when the resistor is disconnected). Look closely at the picture that shows the voltage measurement - the voltmeter red lead is not connected to the resistor.

When the ammeter is connected, the current is much less than it would be if 5.121 volts were across the resistor. So the voltage across the resistor must in fact be much less. This would be clear if there were a 2nd mutlimeter used so the voltage across and current through the resistor were measured simultaneously.

Later in the video, Ben shows that the voltage is not 5.121 V when the resistor and LED are connected to the signal.

enter image description here

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    $\begingroup$ You saved my day!! I really didn't care zooming up at the connections...a really simple solution yet! $\endgroup$
    – user243016
    Commented May 16, 2020 at 19:13
  • $\begingroup$ @MukunthA.G, by the way, I just discovered Ben's YouTube channel this week. I've been watching the series on the breadboard TTL computer from scratch, and I've genuinely enjoyed it. $\endgroup$
    – Alfred Centauri
    Commented May 16, 2020 at 22:09
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Either a bad resistor (value is not 220 ohms) or series resistance in the power supply. He needs to measure voltage across the resistor when signal is applied to find out which is the case.

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  • $\begingroup$ it is clearly mentioned that it is a fine resistor, (he actually measures the resistance in one of the video), and if it is a series resistance, so what?? Ohms law should be applicable right? $\endgroup$
    – user243016
    Commented May 16, 2020 at 17:38
  • $\begingroup$ and indeed he measures the voltage across the resistor when the signal is applied $\endgroup$
    – user243016
    Commented May 16, 2020 at 17:39

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