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So in this case, you think the answer is P = infinity and therefore infinity x zero = infinity and the battery or wire or both just vaporise instantly.
To use LT Spice, you need to set up the resistor as a parameter, and use .setp param... to diminishing resistance. The problem is now that you need to set up an equation for calculating power ie i(R)*i(R)*R. Problem here is, LTspice doesn't conisder R as a tracable quanitity, so the equation doesn't work. The only way I can think is to plot I and then do the power calculation by hand.
PS: I did a few hand calculations and noted that the power rose with diminishing resistance ( approaching zero ) I expect the power to rise monotonically. Thus I conclude that power at zero resistance is non-zero.
I was just wondering something about a simple circuit, that is when you short circuit a battery. You can show that the battery or short circuiting wire will fail by using the equation for power.
What is infinity x 0??? who wins?? Im guessing infinity because the battery will explode and it won't explode if there is 0 watts but how do you prove this? I was always told that anything multiplied by zero = zero and anything multiplied by infinity = infinity but I was never told what infinity x zero = equals.
This problem does not deserve the attention it is getting. There is no circular reasoning, ∞*0 is easily found in this situation, and the calculated power is ∞, not zero.
First of all, forget about a finite battery. Let us assume a ideal voltage source, so that we have a infinite amount of power to draw on.
P = I²R = ∞*0, right? So, we evaluate the term that goes to ∞*0 as R --> 0
P = I²*R = (V²/R²)*R = ∞*0 as R --> 0
So, after cancelling out the terms, P = i²*R = V²/R = ∞ as R --> 0. Therefore, in this case, ∞*0 = ∞ as R --> 0.
In conclusion, shorting out an ideal voltage source causes a infinite amount of power to be drawn from the source.
1) As r goes to 0, i goes to infinity. v = i*r, so the voltage across the resistor = infinity * 0.
But we know the voltage = 1.5 for the ideal battery. Hence infinity*0=1.5 !!
2) As i increases relentlessly it will soon require all the electrons available in the circuit to sustain it. Beyond that point, the immense electromagnetic field due to the current will draw in further electrons from the surroundings. Eventually, it will have drawn in all the electrons in the universe (assumed to be a finite number). Beyond that, the universe will implode and disappear, along with the conundrum.
2) As i increases relentlessly it will soon require all the electrons available in the circuit to sustain it. Beyond that point, the immense electromagnetic field due to the current will draw in further electrons from the surroundings. Eventually, it will have drawn in all the electrons in the universe (assumed to be a finite number). Beyond that, the universe will implode and disappear, along with the conundrum.
The only reasonable reaction to this being, "Don't you dare short-circuit that AA cell with a piece of wire! We don't want to die! We don't want to die!".
1) As r goes to 0, i goes to infinity. v = i*r, so the voltage across the resistor = infinity * 0.
But we know the voltage = 1.5 for the ideal battery. Hence infinity*0=1.5 !!
Yes, depending on the rate at which infinities and infinitesimals (I&I) approach their limit, operations involving division and multiplication with I&I sometimes produce finite results. L'Hospital's method takes advantage of that fact.
1) As r goes to 0, i goes to infinity. v = i*r, so the voltage across the resistor = infinity * 0.
But we know the voltage = 1.5 for the ideal battery. Hence infinity*0=1.5 !!
I came to this same conclusion that infinity x zero = 1.5 V, but if you change the battery to 9 V the same is still true, in fact infinity x zero = any number you can think of. But I can't get round the fact that zero is nothing, so no matter how many zeros you have you should still end up with nothing, even infinite zeros, because if infinite zeros equals anything other than zero then surely you can't have really started with zero in the first place.
But I can't get round the fact that zero is nothing, so no matter how many zeros you have you should still end up with nothing, even infinite zeros, because if infinite zeros equals anything other than zero then surely you can't have really started with zero in the first place.
It makes sense that power tends towards infinity as R tends towards zero and i tends towards infinity, but when you have zero x infinity, it's actually an unanswerable question.
This is not an electronics problem really. It is a math problem. The simple answer is just undefined. infinity * 0 is just undefined. Math helps us understand real world situations, but there is a limit to applying it to all situations. The higher the current, the faster the battery will be drained. That's all there is to it.
You can keep thinking forever about how 1/0 equals infinity, but you will never get anywhere. You see, one apple divided by 1 ( distributed among one person) is one. Mathematically, if you divide one apple by half ( 1/0.5), you get two. But there is nothing as half person.....yet mathematically it is perfectly correct.
You're close, sram, to bringing this discussion to its natural conclusion. Not only is this a math problem; it's a higher mathematical problem. Like way-out-there theoretical math, not the everyday math we use to design and analyze electronic circuits. So this discussion could go 'round and 'round for years without getting anywhere (essentially what they call "mental masturbation").
So we can continue to cluelessly debate things like "infinity times zero", or we can conclude that, in theory at least, power in the shorted-cell circuit does indeed approach infinity.
Or someone could go the fun route and short out a sacrificial AA cell through an ammeter and report back what actually happens.
There is no such thing as an ideal battery. A battery always has a theoretical resistance, or it would not be a battery. A voltage source with no theoretical resistance is an ideal voltage source.
I came to this same conclusion that infinity x zero = 1.5 V, but if you change the battery to 9 V the same is still true, in fact infinity x zero = any number you can think of.
I believe I explained in post #29 why ∞*0 = V, where V can be any voltage you can name. Didn't you read it? I believe it had something to do with infinities and infinitesimals (I&I).
But I can't get round the fact that zero is nothing, so no matter how many zeros you have you should still end up with nothing, even infinite zeros, because if infinite zeros equals anything other than zero then surely you can't have really started with zero in the first place.
How does or how can multiple zeroes or infinite zeroes differ from a single zero? Look, you have to evaluate the variables that converge to I&I so as to get the true value of the term. That is what I did and that is what L'Hospital did over three hundred years ago.
All aspects and principles of electronics are important in various degrees. But this problem is one of the lesser important ones because it is easily to conceive and calculate. We should not be devoting so much time and bandwidth to resolve it.
Anyone who thinks there is too much bandwidth being devoted to this subject should simply stop posting on it, and save the bandwidth his posts are using. The rest of us should take all the time and bandwidth we need to fully understand it.
The result we've been discussing can by shown with very simple math.
LIM I^2*R
R-> 0
=
LIM (V/R)^2/(1/R)
R -> 0
By L'Hospital this is the same as
LIM d/dr(V/R)^2/d/dr(1/R)
R -> 0
which gives:
Lim 2V/R = ∞
R -> 0
or if we choose...
P= V^2/R, and since we know V is the applied voltage ( because we've all done our homework and proved it ) then
Anyone who thinks there is too much bandwidth being devoted to this subject should simply stop posting on it, and save the bandwidth his posts are using. The rest of us should take all the time and bandwidth we need to fully understand it.
The result we've been discussing can by shown with very simple math.
LIM I^2*R
R-> 0
=
LIM (V/R)^2/(1/R)
R -> 0
By L'Hospital this is the same as
LIM d/dr(V/R)^2/d/dr(1/R)
R -> 0
which gives:
Lim 2V/R = ∞
R -> 0
or if we choose...
P= V^2/R, and since we know V is the applied voltage ( because we've all done our homework and proved it ) then
Brown, it is not like this hasn't been discussed or given so much thought. It is a very old problem. It has been beaten to death. The answer is just like I explained. It is like asking what is the largest number. You can't really answer that because you can always add 1 to it and it will just grow larger. You see my point? Here you should stop thinking math, and start thinking logic.
As for the circuit, you can go and try it. If you have an Ammeter, you can hook it in parallel with a 1.5v battery....it will just be like connecting the ammeter in series with a resistance of zero. Better yet, if you have a scope, you can do the same and make graphs for the voltage and current vs time. Do it and tell us what happens ^_^ ........The result can be predicted.
Too bad the ammeter has an internal resistance though.
Haha. Actually infinity isn't there. You can't really use in equations. Anything times zero is zero....however, I wouldn't go as far as saying infinity times anything is infinity. Infinity is a concept designed to help explain some phenomenon.
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