Now, after all this years, this is the first time I know Ohm's first name (never met him dough), some other authors/researchers/inventors I learned the full name at school but I don't recall any teacher referring to Georg Ohm... we are always learning.

Why a topic on Ohm's law, well, just because I needed some relax from more complicated maters (for instance an inverter gate oscillator that is refusing to start with a recycled 4.5Mhz ceramic filter. It starts with a transistor oscillator configuration!).

Now let's do a simple demonstration about Ohm's law:

Place a 9 Volt power supply feeding a 2.2K Ohm resistor in series with a red led (the led is just to brighten things a little bit)

Now measure the voltage across the resistor. In my case I measured the total voltage applyed

then removed the led voltage.

To get: 8.82 - 1.90 = 6.92 Volt across the resistor, measured directly should give the same result.

So, let's calculate the current passing trough the resistor (and the led because it is in series) by Ohm's law:

I = U / R

That's: current is equal to the voltage divided by the resistance.

Since the resistor is a 2200 Ohm one...

I = 6.92 / 2200

I = 0.00314 Ampere

I = 3.14 mili Ampere

let's then measure the current to confirm our calculation:

Now that's a 60 uA (micro Ampere) difference... maybe Mr. Ohm was wrong although 60 uA in 3140 uA is a very, very, very small difference.

But what would happen if my resistor was a 2150 Ohm, as I measured.

Then the current would be 6.92/2150 = 3.21mA that's rounded to 3.2mA and yes...we can confirm that Mr. Georg Simon Ohm was correct!

The very small difference is due to the precision on the multimeter. That's another subject just by it self...

Thank you Mr. Georg for all the fun!

Incidentally I came to the conclusion that I don't know any LED reference value....normally one just ask a 5mm red led...or green. Who cares!?

## 2 comments:

though it isn't really relevant to measuring resistors...

one of the biggest issues with DMMs (at least) is the huge voltage drop on current range. try crefully repeating the experiment with much higher currents and you'll see. even $200 Fluke WILL suck, only a bit less. Dave L. Jones made the uCurrent device to rectify such problems.

A least the calculation gave the expected results :). Anyhow, and for the sake of simplicity I didn't take in account also the load placed in the circuit by the DMM measuring the voltage, that accounts for something around...30-70 pico Ampere.

You can overcome current drop issues by using a DMM for current and one for voltage simultaneously, that way you don't even need to care by voltage drop on the sense element in the DMM.

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