If V(dropped)=I(resistor)*R then
7 V = 0.5 A * x Ω
x = 7 V / 0.5 A = 14 Ω
It doesn't seem like I calculated corectly. Plus in the ecuation I have only the intensity of the resistor(500mA), what about the intensity of the power source?
and what if I want to lower the intensity. I have:

Power source:
Tension: 12 V
Intensity: 500 mA

Thing I need to power:
Needs 5 V
No more than 100mA

I need to know what resistor should I buy. How many Ω and how many ampers should the resistor have?

 

Hello,

Better look for a voltage regulator like the LM7805 or a switching module.
This will avoid voltage changes when the load changes.

Bertus

 

Almost right. (12V-5V)/100mA = 70Ω
Power rating will be (12V-5V) x 100mA =0.7W
But bear in mind that if the current is LESS than 100mA, then the output voltage will be too high. Much better to use the voltage regulator.

 

Welcome to AAC!

Resistors have two important specifications:
1) resistance in ohms
2) maximum power rating in watts

Operating current and voltage specifications can be determined by applying Ohm's Law and power formulas.
(These are based on max power dissipation and safe operating temperatures. Absolute safe max voltage is an additional specification.)

Using a fixed resistance only works for a fixed load current. A better solution would be to use a voltage regulator such as LM7805.