Learning electronics. But something doesn´t sense with resistor

Thread Starter

Duffman83

Joined Apr 25, 2020
13
Hello.

Sorry if I make mistakes, I am learning english. With the quarantin thing, I have started playin with electronics, but I realise I have no idea about the basics, so I read the articles about Ohm´s law and circuits( Vol. I - Direct Current (DC)) in that web. They are really good and explanatory!

I made an experiment, I have conected a 220 ohms and 3w resistor in an old laptop 19V power supply. My calculations were correct. 0,086 Amps. So, the disipate power must be 0,086x19 = 1,64 watts. But the resistor got very very hot in less than a minute. What is wrong? It is rated for 3 watts.

Thank you.

Best regards
 

OBW0549

Joined Mar 2, 2015
3,411
Probably nothing is wrong; when resistors are run at their maximum rated power, they do get hot. Usually, too hot to touch.
 

Ylli

Joined Nov 13, 2015
955
Nothing is wrong. 3 Watts is the maximum the resistor can dissipate without exceeding it temperature limitations. Those temperature limitations can be rather high (+150°C for example). A 3 watt resistor dissipating 1.64 watts will become quite hot.
 

Thread Starter

Duffman83

Joined Apr 25, 2020
13
Ok, thank you for your responses. I read about voltage divider circuit. I need around 4 or 5 volts to electroplate with cooper one metal piece. I have designed this circuit. I have measured 195 ohms in the electrolite (Salted water). If I conect the the metal piece to anode and the cooper piece on catode it is suposed to flow a current of 0,017 amps. Is that correct?

cir.jpeg
 

Ylli

Joined Nov 13, 2015
955
Voltage dividers don't work too well if you are trying to draw any significant power from them. In your case, if you just want to force 15 mA through the plating bath, a simple series resistor will give you the best results (closest to a constant current source).
R = E/I = 19/0.015 = 1267 ohms.
Your plating bath makes up 195 ohms of that, so the added series resistor would be:
1267 - 195 = 1,072 ohms.

A 1000 ohm (1K) resistor would be close enough. That resistor would dissipate:
P = I²R = 0.015² * 1000 = 0.225 watts. I would suggest a 1 Watt resistor.
 

Thread Starter

Duffman83

Joined Apr 25, 2020
13
Are you working on a time machine?
hahaha, sorry, I need improove both english and electronics.


Voltage dividers don't work too well if you are trying to draw any significant power from them. In your case, if you just want to force 15 mA through the plating bath, a simple series resistor will give you the best results (closest to a constant current source).
R = E/I = 19/0.015 = 1267 ohms.
Your plating bath makes up 195 ohms of that, so the added series resistor would be:
1267 - 195 = 1,072 ohms.

A 1000 ohm (1K) resistor would be close enough. That resistor would dissipate:
P = I²R = 0.015² * 1000 = 0.225 watts. I would suggest a 1 Watt resistor.
Thank you very much. I will do that
 

MisterBill2

Joined Jan 23, 2018
6,105
The important thing to understand about any voltage divider is that every device that has current flowing though it is part of that circuit, and therefore mus be included in the calculation. That is why sometimes the results are not what is expected.
 

Ylli

Joined Nov 13, 2015
955
By the way Duffman83, your original circuit did give you 15 mA through the plating tank, it just was not the most efficient way to do it.
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