I'm taking electronics classes at the moment, and I've noticed something very interesting about a special case in one of the equations.
Capacitive reactance is the measure of how much a capacitor resists the flow of power. Capacitors allow high frequency to pass easily, but tend to block low frequencies. DC power is the lowest frequency possible (0), so it is blocked completely (infinite capacitive reactance).
The formula for capacitive reactance is: X=1/(2(pi)fC)
f is for frequency and C is for capacitance
Let's suppose we have a .05 farad capacitor and DC current (0 frequency)
That would make the equation: X=1/(2(pi) * 0 * .05)
Anything multiplied by 0 = 0 so: X=1/0
Normally we would be stuck at this point, but we already know the answer: X must equal infinity.
So, 1/0=infinity
I know there are problems with it being reversible, but I'm working on that. The best I've been able to work out as a solution to that so far is X/0=(infinity * 0 * infinity)/X
you can't use infinity as a number! 1/0 is not equal to infinity. 1/0 is undefined. if we have 1/a and let a be smaller and smaller 1/a will go towards in the infinity.
Well 1/0 is undefined because the value would be too infinitely large to describe in numbers. But infinity isn't a number so I wouldn't say that 1/0 is infinity.
Other people have already pointed out the "infinity is a number" fallacy, so i won't repeat it. (i will offer the standard alternative i usually do: infinity is not a number, it is a direction.)
Instead, i'm gonna snipe at the carelessness of this "experimentally verified" claim.
Even assuming that one were to actually set up this "experiment" - which doesn't really need to be run, because it is a principle that many devices already use in practice, so we can consider it a fait accompli - it "verifies" nothing.
For starters, there is demented logic of trying to verify a mathematical concept by applying that mathematical concept. In simple terms, it is like saying: "We don't know the ratio of a circle's diameter to its circumference, so we'll check it experimentally. So i make a circle, and measure, and since i already know the answer is π, and that's i seem to get, that means it's proven!" See the problem? ^_^
But the real problem is the impossibility of such an endeavour. In real life, meters have limits. In real life, you would suppose you have a capacitor of 0.050 ± 0.002. When you attach your DC ammeter next to your capacitor and find a current of 0 A, what you are really finding is something more like 0.000 ± 0.002 µA. Small, but not certainly zero. Which means that if you use your voltage drop (let's say 5.000 ± 0.001 V) to find your reactance, you aren't really dividing by 0.
But, more importantly, in real life, there is no such thing a truly 0 Hz (true DC) signal. It is a theoretical idea only, and impossibly to get in reality. Even the most expensive battery drains slowly as it works. And even if you were to somehow set up a system that did not drain, it is impossible to maintain a perfect potential difference when charges are moving - quantum chromodynamics tells us this with certainty... but if you take a moment and realize that charges are little packets of negative charge, and imagine trying to set up a situation where the charges on both sides of the circuit are constant while these little packets keep jumping, you will see that there will be tiny fluctuations that cannot be corrected faster than the speed of light times the length of the circuit.
Be careful with your claims. ^_^ "Experimentally verified" is a bold claim.
Is that formula correct in extreme situations such as this one? If yes: are your instruments correct in such situations?
I was thinking about this today, and I thought of another "proof" for you. If you're going to try and prove that x/0 is infinity (even though infinity isn't a number; I prefer to say that x/0 is too infinitely large to be defined by numbers, which is why it's undefined), then try doing this instead.
Integral between pi/2 and p/4 of sec^2(x)...
When you take the antiderivative of sec^2(x), you'll get tan(x). After plugging in the endpoints, you'll see that you'll get (undefined - 1)... Graphing this out, you'll see that sec^2(x) has a vertical assympote at pi/2... so obviously the area under the curve will be infinity...
sec x is such an American function. At my university we just call it 1/cos x. Anyway, sec˛ x is the derivative of tan x, probably the only reason why it's defined...
In the example above: cos(0.5pi) = 0 so sec(0.5pi) divides by zero. This means that at exactly that point of the graph the function is undefined, but when approaching it you approach infinity.
Whether the area will be infinity is another question, because although the y component in the area near the asymptote is extremely large, the x component is extremely small.
You've proved the theory using, um, theory. There is always going to be a small leakage current through the capacitor and so it's not fully blocked. But if you're doing electronics then that sounds like you're more likely to be taught the engineers way of thinking which is to approximate, simplify, linearise, simplify a bit more and then it works easy (but add in a factor of safety just in case) rather than the scientist/mathematician thinking which requires no such simplifications but rather needs to expand on it in order to fully explain it. So an engineer is quite happy to accept that there is a current of practically zero which means a practically infinite reactance- engineers don't need to prove it, just use it (unless they're at uni in which case all they do is prove things).
| Arnie wrote: |
sec x is such an American function. At my university we just call it 1/cos x. Anyway, sec˛ x is the derivative of tan x, probably the only reason why it's defined...
In the example above: cos(0.5pi) = 0 so sec(0.5pi) divides by zero. This means that at exactly that point of the graph the function is undefined, but when approaching it you approach infinity.
Whether the area will be infinity is another question, because although the y component in the area near the asymptote is extremely large, the x component is extremely small. |
I'm sorry; we use secx at my school (which is in America) and then when I talk to other people or right problems out that other people see, I try to use 1/cosx (because that's universal; many places use secx, but many places - like your country - do not). But sometimes I forget when I go back and forth and simply use secx out of habit. I was also unsure for a while rather it would be infinity or not, but after thinking about it for a while and going over with several professors/colleagues, I came to the conclusion that it must be infinity. After studying the graph and seeing the reaction of multiple computer programs, it seems like it will be infinity despite the small x component.x/0=infinity ????? 
Then i think you will have to correct everyone out there...!!!
| ocalhoun wrote: |
Capacitive reactance is the measure of how much a capacitor resists the flow of power. Capacitors allow high frequency to pass easily, but tend to block low frequencies. DC power is the lowest frequency possible (0), so it is blocked completely (infinite capacitive reactance).
|
You lost me here... I thought capacitors are DC devices and only allow (vs block) DC current.
PS are you a 13 year old and reviewing some NASA publications also 
?
http://www.freerepublic.com/focus/f-news/2002114/postsNo - that's a rather oversimplistic view of the capacitor (well, actually plain wrong - the simple view is the other way around - capacitors block DC).
Wiki has a reasonable article on the matter
http://en.wikipedia.org/wiki/Capacitor hmmm ... I've built circuits for years, they block current in that they store similar to a battery, only under certain circumstances do they actually block DC. Look on any circut board, those guys are not there to block DC. For ac circuits I use them to bring down the spikes or curves in the voltage or signal.
| czc587 wrote: |
| hmmm ... I've built circuits for years, they block current in that they store similar to a battery, only under certain circumstances do they actually block DC. Look on any circut board, those guys are not there to block DC. For ac circuits I use them to bring down the spikes or curves in the voltage or signal. |
That'll be why I said it was a simplistic view then
I've also built circuits for years, and frequently built both low pass and high pass filters, both of which use...yep, capacitors.
The fact that capacitors block DC (in ALL circumstances) can be easily seen from the basic current derivative of a capacitor.
(where i(t) is current at time t, V voltage, C capacitance).
Note that for direct current dv/dt=0 so the current passed (regardless of capacitance) is 0.
In DC circuits their primary use is for temporary storage of current, or power shunts. Their main use is in AC circuitry for filters and tuning circuits, and in mixed circuits they act as couplers/decouplers between DC and AC.1/0=Infinity then ------- 0*Infinity=1 -----------. I didnt know this notable equation!!!
I always like the following argument:
1/2 means you take one thing and divide it into 2 parts
1/0 means you take one thing and divide it into no parts - what do you have then? When you phrase it like that you realise how meaningless the question is.
You're looking at the left side of the graph and seeing that the red line at x=zero must be infinity high.
Well, now look at the "right side" of zero. You get negative infinity.. Hm... that's why we don't say that 1/0 is infinity: it might be, for example, negative infinity.
