Thus, the voltage must fall. In this extreme example, the voltage would fall to zero.
you are incorrect.
Nope. I'm exactly correct.
current and resistance are directly proportional to voltage.
V (voltage)=I (current) x R (resistance)
Right.
All three are related.
Now plug in some numbers. Remember!... you don't have unlimited current available.
Here you go:
Suppose you have a 12V supply that can provide 30 amps.
If you load it with .8 ohms, what it be supplying? 15 amps, right?
If you load it with .4 ohms, now it is supplying 30 amps.
What happens if you load it with .2 ohms?
I=E/R , so would current be 60 amps? Of course not - the ps can't do it.
So, we now we have R and I (fixed at 30 max). E=I*R , thus voltage is
6V.
The voltage must fall.
You were correct in quoting Ohm's Law, and it is the reason the voltage will fall.
if what you are saying is true then, according to this mathematical formula, electricity would cease to flow once a short circuit is introduced.
In theory, that is exactly correct. And it is not what I'm saying, it is what Ohm said.
Try it. Plug numbers into Ohm's Law.
Pick a voltage. 12v.
0 ohms - that's your theorectical short circuit ohm value
I=ER
I= 12 * 0
I= 0
this would mean it would be impossible to destroy a battery by shorting the two terminals.
Because there would no longer be any potential difference in the two terminals.
But.... that's not what really happens when you "short" them.
If you have a defective power supply.... suppose for some reason it can only supply 10 watts and it is a 300 watt supply, then it doesn't take much of a load before the voltage begins to fall off. However, with zero load, the voltage will be just fine.
the voltage will not drop unless there is a problem
Bingo!
Now you've got it. The voltage may drop if there is a problem.
And this is not what you said earlier:
"the voltage at the supply will stay the same regardless if a load is placed on it or not." , that I pointed out was incorrect.
with the voltage regulators.
i will go more in depth later, but i am suffering from the flu right now and i can not hold a thought for more then 10 seconds.
Ah. An example of a voltage divider.
that was not an example of a voltage divider, it did not have a voltage out, that was an example of a simple series circuit. one must have the resistor there to drop the voltage from 5V to roughly 1.5V, the voltage most common LEDs use. i also included 2 loads
??
Each will have its own voltage drop across it. Simply measure that, and you'll see. That's a voltage divider.
to show you that voltage drops happen after the load.
I don't know what you mean here... voltage drops happen
across loads.
But that is ok.... you seem to want to argue with me, and I'm not interested in that. You made a rather blanket statement that was not correct, and in the interest of the forum readers, and also to do you a bit of a favor, I pointed it out. With Ohm's Law that you are obviously familiar with, and some consideration about both the theoretical and the real, you'll grasp it.
