Not telling me anything I don't already know... but v(t)/i(t) proves when a circuit exhibits reactive properties, the ohms value from instant to instant is not constant. Additionally, it even goes negative when power is negative.
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- Thread starter rattus
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Smart, that is absolute, utter nonsense. This "ohms value" is not defined anywhere in the literature. Neither does it prove anything. You should not confuse the newbies with your unproven theories.
Surely enough it is. For the sake of discussion I'll pretend to take Besoeker's disposition of there being no real or reactive power involved with intantaneous measure of voltage and current. As such, there would be no reactive impedance, no alternating current, etc, and thus voltage and current measure of an instant can only yield two derived measures through the equations of P=EI and E=IR... and there's the R.
Oh wait a minute... I forgot we are not permitted to use chapter one :roll:
Cold Fusion
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uhhhh .... that is called "integral calculus" . Which is generally considered a little more than, "just a little trig and arithmetic" And if you want the reactive component, then there is this pesky phase angle business. Your statement is more false than true.
I'm surprised, your math background appears pretty heavy.
cf
Cold Fusion
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Encrypted for Dave's eyes only.
Dave -
Is rattus correct? He can read my encrypted posts?
Oh no the shame of it.
cf
What now?
What now?
Now why would one need calculus to evaluate a trig function? And, why is there a phase angle? The value of a sine function does not carry a phase angle--just a numeric value according to my obsolete trig tables.
What now?
Now why would one need calculus to evaluate a trig function? And, why is there a phase angle? The value of a sine function does not carry a phase angle--just a numeric value according to my obsolete trig tables.
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gar
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100514-2142 EST
rattus:
In your first post you provided no limitations. So I choose to consider the case of a fixed resistor, R.
This real world resistor has a resistance value of R independent of what ever voltage is applied to it, including 0 volts.
If the resistor is 100 ohms and we apply different voltages to it and measure the current, then the result is observed to be v(t)/i(t). This is true even at 0 voltage because in the limit as we approach 0 for the divisor we find that the ratio still remains the same 100 ohms.
You can divide by 0 if you know how the equation solution varies as you approach 0 as the divisor.
.
rattus:
In your first post you provided no limitations. So I choose to consider the case of a fixed resistor, R.
This real world resistor has a resistance value of R independent of what ever voltage is applied to it, including 0 volts.
If the resistor is 100 ohms and we apply different voltages to it and measure the current, then the result is observed to be v(t)/i(t). This is true even at 0 voltage because in the limit as we approach 0 for the divisor we find that the ratio still remains the same 100 ohms.
You can divide by 0 if you know how the equation solution varies as you approach 0 as the divisor.
.
Try posting a little faster. If you can make them instantaneous, Rattus will still see them, but he won't believe they exist
Its all in fun rattus
Cold Fusion
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Some would think that the phrase evaluate "sin(wt) and sin(wt + phi) over a cylce" implies an RMS calculation, which is an integral.
That would be Euler.
I would have thought you had one of new fangled taylor series calculated sets. According to my history teacher, they were developed some time in the 1400s? So, yours are older than that?
cf
Some would think that the phrase evaluate "sin(wt) and sin(wt + phi) over a cylce" implies an RMS calculation, which is an integral.
That would be Euler.
I would have thought you had one of new fangled taylor series calculated sets. According to my history teacher, they were developed some time in the 1400s? So, yours are older than that?
cf
cf, you must be from another planet because you seem not to understand English, or maybe you are just stonewalling.
Cold Fusion
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Oh my goodness - I believe I am being put in my place as an phillistine that should know better than to challange the great RATTUS. I'll risk the lightning strike with one more response.
Be assured I have no expectation of convincing you - I'm quite sure you already understand. This is just for the others listening in.
So, nope, you are just dead wrong. I know that is hard for you to believe with all of your years of experience. But it is true.
Evaluation of a sinusiod over one period is an integration. It's even defined. You know that - I know you know that. I'd say most everyone past high school trig knows that.
Your response with a personal attack is indicative you have taken an unjustifiable position - and you got caught.
I refuse to respond in kind.
You have a good day
cf
Look Ma, no calculus:
Look Ma, no calculus:
Since some have experienced trouble with this question, I will demonstrate:
Let,
v(t) = Vm*sin(wt)
i(t) = Im*sin(wt ?pi/6)
Now, for example, let us evaluate the two functions at wt = (pi/6):
v (pi/6) = Vm*sin(pi/6) = Vm*0.5
i (pi/6) = Im * sin(pi/6 ? pi/6) = Im*0.0
Then divide the results,
v(pi/6)/I(pi/6) = 0.5/0.0 = infinity
Evaluation of additional points will show that the ratio swings between +/- infinity.
Now what good is this ratio which some call z(t)?
Look Ma, no calculus:
Since some have experienced trouble with this question, I will demonstrate:
Let,
v(t) = Vm*sin(wt)
i(t) = Im*sin(wt ?pi/6)
Now, for example, let us evaluate the two functions at wt = (pi/6):
v (pi/6) = Vm*sin(pi/6) = Vm*0.5
i (pi/6) = Im * sin(pi/6 ? pi/6) = Im*0.0
Then divide the results,
v(pi/6)/I(pi/6) = 0.5/0.0 = infinity
Evaluation of additional points will show that the ratio swings between +/- infinity.
Now what good is this ratio which some call z(t)?
Or you could realize that it takes both impedence and reactance make the current shift in phase. So the problem is easily solved for v(t)/i(t) equaling 0.866 ohms of resistance in series with 0.5 ohms of capacitave reactance.
Steve
But Steve, there is no way that the ratio of v(t)/i(t) could be 0.866 -j0.5 because the ratio, v(t)/i(t), at any point in time, is a single real number. Furthermore, the ratio would have to be constant before it could be called "impedance", and it is not.
Simply put, impedance and reactance are undefined for instantaneous equations. You must use RMS values in a steady state analysis.
drbond24
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Hello everyone, I'm back! I don't look at this stuff over the weekend, but I've had fun catching up this morning. :grin:
CF, I appreciate the encrypted messages you've sent me. They have been helpful.
I think the problem with this thread is that the limitations were not clearly spelled out in the OP. If the professor had just said "Does the ratio v(t) / i(t) mean anything if you don't use any math more advanced than they teach in kindergarten," we'd have all got the answer he was looking for right away. I could have just showed it to my daughter (2nd reference to her, so I should say she's 5 and just finishing kindergarten) and she'd have told me it looked like a load of crap and the matter would have been settled right then and there.
However, a few of us here have actually had calculus and differential equations, so even though an individual's math knowledge may seem like a superpower to the rest of the average outside world, on here there is always someone smarter unless, of course, you're me. :grin::roll: haha
CF, I appreciate the encrypted messages you've sent me. They have been helpful.
I think the problem with this thread is that the limitations were not clearly spelled out in the OP. If the professor had just said "Does the ratio v(t) / i(t) mean anything if you don't use any math more advanced than they teach in kindergarten," we'd have all got the answer he was looking for right away.
I could have just showed it to my daughter (2nd reference to her, so I should say she's 5 and just finishing kindergarten) and she'd have told me it looked like a load of crap and the matter would have been settled right then and there.However, a few of us here have actually had calculus and differential equations, so even though an individual's math knowledge may seem like a superpower to the rest of the average outside world, on here there is always someone smarter unless, of course, you're me. :grin:
:roll: haha
mivey
Senior Member
It is used in transient analysis and is the formula for transient impedance.
mivey
Senior Member
As a fellow utility engineer, I'm sure you are aware it is of interest to us.Of course it has meaning. All expressions have meaning. The matter of whether its meaning is of engineering interest is another question.
Not true.
Nope
For some applications that is true.
There are many other posts I could respond to but I'll just skip here to the end and say:
Rattus, there is a field of study dealing with transient impedance so your blanket statement is just not completely true.
mivey
Senior Member
Oops, wrong charlie. But a compliment to both, I'm sure.
Yes, but:
Yes, but:
Agreed, but we have for the most part limited our discussion to sinusoids, and is this "transient impedance" vary with time?
Sounds like it is more like Zo of a xmission line??
Yes, but:
Agreed, but we have for the most part limited our discussion to sinusoids, and is this "transient impedance" vary with time?
Sounds like it is more like Zo of a xmission line??
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