Okay, so here are my questions:
By having X volts running trough a conductor, if the conductor resistance is Y, then the current across that conductor will be X/Y according to Ohms law, right?
Will the current be ALWAYS X/Y NO matter of the load OR the load will suck the current he needs? I mean will the resistance Y set the current flow if the volts X dont change or the load sets it?
I have 12V power supply. If I connect 2ohm 100W power resistor will then the current be 6A and the heat losses 72W?
You don't have current across a conductor, you have current flowing through a conductor.
Voltage across a load divided by the load resistance equals the current flowing through the load.
V/R=I
P=V*I
Nothing more complicated that this.
The math is your job.
JMD1:
I mean if I have 12V battery and I connect 2R resistor (and there is no other resistance in the circuit) will then the battery be forced to give 6A?
If you have a 2R resistor connected directly across a 12V battery with nothing else in the circuit then the current will be approximately 6A.
But if you have the resistor and a "load" like you mentioned in your first post, maybe a motor or high power LEDs or something, then even though they are not actual resistors they will drop some of the 12V so they will reduce the current.
JMD1:
Okay, so here are my questions:
By having X volts running trough a conductor, if the conductor resistance is Y, then the current across that conductor will be X/Y according to Ohms law, right?
Will the current be ALWAYS X/Y NO matter of the load OR the load will suck the current he needs? I mean will the resistance Y set the current flow if the volts X dont change or the load sets it?
Its true always because that is how resistance is defined: R = V/I. For metallic conductors and various
other materials R is independent of I, such materials are said to obey Ohm's law (which is that V is
linearly proportional to I)
In many situations and devices Ohm's law doesn't hold (capacitor, inductor, a vacuum, a super conductor, a non-uniform
semi-conductor, a quantum tunnelling device. But R is always defined as V / I (for non-ohmic situations
R is not constant as I varies - ie its a function of I or time, etc)
When we move from DC conditions to AC R is replaced by Z (impedance) which generalizes to include phase
information.
And there is another notion of resistance called dynamic resistance, which is defined as R = dV/dI, ie
the slope of the V-I curve - this is useful for analysing small ac signals flowing through non-linear devices
like diodes and transistors. You sometimes see this quoted for zener diodes for instance.
Not at all. The battery will source as much current as it can. R2 demands 6A. If the battery is capable of delivering 6A, congrats, you just made a coffee cup warmer. If that 6A demand is more than the battery can source, it appears as a short to the battery. The current will be at max while the voltage drops. Smoke, fire, brimstone follow.
Maybe not brimstone, but is is important to know the capabilities of the psu and the load you are placing on it. Ohms law will help you with that
tinman13kup:
Not at all. The battery will source as much current as it can. R2 demands 6A. If the battery is capable of delivering 6A, congrats, you just made a coffee cup warmer. If that 6A demand is more than the battery can source, it appears as a short to the battery. The current will be at max while the voltage drops. Smoke, fire, brimstone follow.
Maybe not brimstone, but is is important to know the capabilities of the psu and the load you are placing on it. Ohms law will help you with that
But how am I supposed to find out what are the capabilities of the battery? If I have 2000 mAh, then I will make it give me 6A for 20 mins. Under what circumstances the smoke and fire you mentioned may occur?
I think what @tinman13kup is saying is that not all batteries are able to supply 6A e.g. if you made your 12V 2000mAh battery out of 10 x 1.2V AA batteries it probably couldn't.
If you put a 2R resistor across a battery that is not capable of 6A then you won't get 6A. Instead you will damage the battery.
JMD1:
But how am I supposed to find out what are the capabilities of the battery? If I have 2000 mAh, then I will make it give me 6A for 20 mins. Under what circumstances the smoke and fire you mentioned may occur?
The datasheet for the battery will have discharge curves at various rates, and possibly internal
resistance and max-charge/discharge ratings.
Just looked at my assortment of "normal" wall adaptors and none of them will supply anywhere near 6A. Most are around 1-1.5A max. But the good thing is that in every case the amount of current they can supply is written on them. It's usually just after the writing that tells you about the 12V.
And random 9V batteries will supply random amounts of current, many of them much less than 4.5A. E.g. the most common 9V batteries are those rectangular things you get in smoke detectors and you'll be lucky to get even 1A out of most of those.
Round here we're mostly engineer types. We work with specifications and datasheets so we have a good chance of getting things right. We try not to work with random things and guesswork because that usually gets things badly wrong.
Batteries in parallel combine current, while retaining the same voltage.
I have a diesel truck. It takes , say 800 cranking amps to spin the motor. 1 battery can provide 450CCA. I have 2 batteries in parallel, which provide 900 CCA @12Vdc.*
The numbers are just guessing. I don't recall the exact amps required, nor the CCA of the batteries, but the reasoning is correct.
Trying to source more current from a battery than it can produce creates problems, like heat, and lots of it. Enough to destroy the battery, blow up a battery, or catch fire. That truck I have, it melted the lead battery connector. Yes, it looked like someone melted solder all over the place. That was probably due to some oxidation on the terminal, which acts as a resistor.
Know what your power source can provide, and don't push the limits on it unless you feel lucky. Sooner or later, it will turn and bite you.