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/bazaznanja/e ... zakon.html
I want to build an soldering iron
I want to build an soldering iron
 Attachments

 unimer.tap
 Program
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 unimer.tap
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 screenshot
 SCREEN1 copy.jpg (25.37 KiB) Viewed 3697 times
Last edited by Brana on Mon Nov 26, 2018 1:41 am, edited 7 times in total.
Re: I want to build an soldering iron
It IS correct!
According to HIS calculator at 02:00
According to HIS calculator at 02:00
Re: I want to build an soldering iron
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 !
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 !