I've noticed that Ohm's law is used like everywhere!!! But it's hard for me to understand how to use it...
General writing: V=Z*I I being the current.
Z being the Impedance (R, 1/JwC, jwL "Resistive, Capacitive, Inductive").
if i have an LED which runs on 1.5 V but i don't have 1.5 V battery, i only have 12V battery, and if i attach that LED to 12V battery it would burn emediatly, so i want to know how strong the resistor must be to resist those 10.5 volts
There is some disambiguation needed here:
*The PN junction in the LED needs a certain direct potential (anode cathode) VAK called threshold tension Vs or V0 in order to be conductive (ID<>0)
*The LED will cause a tension drop of that amount.
*The LED has a certain resistance r, that is small.
The reason the LED would burn isn't "because" it is 12 volts, the reason it would burn is because the CURRENT circulating through the LED (ID) is so huge, it will cause an avalanche, a snow-ball effect. Back to Ohms law, once you pass the threshold tension, V=r*I so I=V/r .. This is big, as r is small.
If r=10ohms and V=12 Volts, you get I=1.2 Ampères. That's a lot for our poor LED.
The point is, it's not the tension that hurts, it's the current that hurts.. So, you want to put a Protection Resistor not to "resist a tension" because the tension will still be 12 volts, but to "limit" the current circulating in the circuit.
Without the Protection Resistor (Rp):
I=(V-Vs)/r (Vs the tension drop in the LED, r or the diode is too small, I will be huge
)
With the Protection Resistor:
I=(V-Vs)/(r+Rp) (I diminishes).
and how to know how many Volts equals to 1 Ampere, or how many Amperes equals to 1 Volt.
I wish you could help me with this, because if i wont unerstand this i would be stuck int this level forever... Thank You
Here again, let's get back to Ohm's Law:
V=Z*I , Z in Ohms, V in Volts, I in Ampères. So {Volt}=={Ohm}*{Ampère}
So, {Ohm}=={Volt/Ampère}.
It is an equation with three parameters, which means that you ABSOLUTELY NEED TWO KNOWNS to get the THIRD.
Let's say you want a Current I of 1 Ampère, you can get it with
10Volts/10Ohms=1 Ampère. (The 10 Ohms can be Two 5 Ohms Resistors in Series, or Two 20 Ohms Resistors in Parallel, whatever combination you wish)
2KVolts/2KOhms=1 Ampère. Whatever you want and is smart according to your context.
So if you want a current of 20 mA (milli-Ampère = (1/1000) Ampère) You'll have:
V=Vs+(Rp+r)*I ... Let's calculate the value of the Protection Resistance to have that current...
(Rp+r)*I= V-Vs
Rp+r=(V-Vs)/I
Rp=(V-Vs)/I - r
Suppose Vs=1.5 Volts
V= 12 Volts
r= 10 Ohms
Rp= (12-1.5) Volts/(20*1/1000)Ampères - 10 Ohms
Remember, Volt/Ampère==Ohm
Rp= 525 Ohms - 10 Ohms = 515 Ohms .. (Example for educational purposes only ..)
If there is anything you didn't get, feel free. (If so, would you please tell which level you're in to get more pertinent explanations)