forum.defence-force.org

Yes, I want to build an soldering iron...
...And the first step is to - turn On the Euphoric

So I created (first) the following program:
It is on Serbian (sorry about that!) but it is really so simple stuff.

- Enter ANY desired two values on the left
(Use Up/Down arrow keys to move, Enter to - enter the value, R to restart).
- As soon as at least two values are entered, program will then calculate all other needed values on the right side.

My soldering Iron should work on 12 Volts
And I would like to have 30 Watts of power.

So (on the left side) I enter 12 Volts And 30 Watts.

And I get 4,8 Ohms impedance for the soldering iron needed,
With the power adapter of 2,5 Ampers strength

CAN ANYONE CONFIRM THIS IS CORRECT?
Really?

Source for building the program was Wikipedia and
https://www.automatika.rs/baza-znanja/e ... zakon.html

It IS correct!
According to HIS calculator at 02:00

Well Brana,
I'm not very skilled in electricity or in electronics, but according to the basic Ohm law, yes, your calculations are correct for your example:

Known values:
P = 30 W
U = 12 V

Using the P = U x I equation we can then compute I = P / U = 30 / 12 = 2.5 A

Then using the U = R x I equation we can finally compute R = U / I = 12 / 2.5 = 4.8 Ohm

All good !