Something Different:

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kingpb

kingpb

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rattus said:
What if the load is reactive? What then?
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The reactive nature of the circuit would cause the phase shift between the curent and the voltage. It doesn't change the fqct that the ratio of v(t)/i(t) is still meaningless for power engineering applications.

BTW: I didn't say nonsense, which would imply language, conduct, or an idea that is absurd or contrary to good sense, I said it was meaningless for the general practicality of power engineering. Now v(t) x i(t) is a different story.:D
 
Smart $

Smart $

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drbond24 said:
Then v(t) / i(t) = Z, but you've already shot that one down. I'm going to go back to patiently waiting for you to give up the answer. :)
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If V/I = Z, then v(t)/i(t) = z(t)? ;)

Other than making interesting lines on a chart of instantaneous values and showing that impedance for reactive circuits is an average value of ohms rather than a constant value of ohms, it is rather meaningless.
 
Hameedulla-Ekhlas

Hameedulla-Ekhlas

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rattus said:
Who thinks that the expression,

v(t)/i(t)

has any meaning? If so, what meaning?
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Ok for something different question something different answer

v(t)/i(t) =(I[Vcos∅ { 1-cos⁡(2wt) }- Vsin∅sin(2wt) ]^2)/(2√(2 ) Psin(wt-∅)sin⁡(wt))


Now who can go reverse to prove v(t)/i(t) step by step ?
 
G

gar

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drbond24:

I presented the R answer near the very beginning. There are many answers to rattus's broad question.

I will present another one. Instantaneous resistance for a name. And a particular case is a motor driven variable resistance from 0 to R1 with an added fixed resistor R2 in series with R1. Apply any desired voltage waveform you desire and the current will be i(t) = v(t)/r(t). If you know the voltage and current at a given instant you can calculate the resistance at that instant and in turn the position of the variable resistor.

.
 
Hameedulla-Ekhlas

Hameedulla-Ekhlas

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Ok for something different question something different answer

v(t)/i(t) =(I[Vcos∅ { 1-cos⁡(2wt) }- Vsin∅sin(2wt) ]^2)/(2√(2 ) Psin(wt-∅)sin⁡(wt))

let me clear the question a little bit

b3x7xf.png


Now who can go reverse to prove v(t)/i(t) step by step ?
 
steve66

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Smart $ said:
If V/I = Z, then v(t)/i(t) = z(t)? ;)

Other than making interesting lines on a chart of instantaneous values and showing that impedance for reactive circuits is an average value of ohms rather than a constant value of ohms, it is rather meaningless.
Click to expand...

Yes, v(t)/i(t) = z(t).

What objection does everyone have to that?

Of course, z doesn't really change with time. Its a constant magnitude, with a phase angle.

Steve
 
steve66

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rattus said:
We are in Chapter 1. Impedance is covered in Chapter 2 or maybe Chapter 10. I forget.
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So Chapter 1 is no longer true after we've read chapter 2???

I'm with the sponge. If you have a point to make, please do so.

This is starting to look a lot like the "just for fun" thread you started. After 10 pages, I would have thought you would have tried to make some kind of point by now.


Steve
 
R

rattus

Senior Member
Objection!

Objection!

steve66 said:
Yes, v(t)/i(t) = z(t).

What objection does everyone have to that?

Of course, z doesn't really change with time. Its a constant magnitude, with a phase angle.

Steve
Click to expand...

The point is that the ratio, v(t)/i(t) varies from zero to infinity in a typical setting and is of no use in circuit analysis. Impedance, a constant, is defined for steady state analysis--phasors if you will. We are not into phasors in this discussion.

Utter nonsense!
 
Smart $

Smart $

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steve66 said:
Yes, v(t)/i(t) = z(t).

What objection does everyone have to that?

Of course, z doesn't really change with time. Its a constant magnitude, with a phase angle.

Steve
Click to expand...
Did you mean "z" or "Z"? Because v(t)/i(t), aka little z as in z(t), plotted vs. time is tangential.
 
Smart $

Smart $

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rattus said:
My out of print texts written by long dead profs do not mention z(t). Is this something new?
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"Something new, something old, something borrowed ..." Ohh, that's something else entirely :roll:

I don't know... it just follows that if V/I = Z that v(t)/i(t) = z(t)
 
R

rattus

Senior Member
Smart $ said:
"Something new, something old, something borrowed ..." Ohh, that's something else entirely :roll:

I don't know... it just follows that if V/I = Z that v(t)/i(t) = z(t)
Click to expand...

Think about it Smart. V and I are constants, therefore Z is a constant. Not so with v(t) and i(t). Does not follow.
 
G

gar

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Ann Arbor, Michigan
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EE
100512-2201 EST

For those that are concerned about dividing by zero and have not had differential calculus the following is background.

If y = A*x you have an equation for a straight line. The slope of this line is at any desired point the ratio of a small incremental change in y divided by the small incremental change in x causing the y change. As you make the increment in x smaller and smaller it approaches 0.

Now you need to know how delta y / delta x changes as x approaches 0. It turns out in this case that the ratio remains constant and equal to A for any value of x including 0. So indeed you can divide by 0 in some cases and it won't be infinity for the result.

.
 
Smart $

Smart $

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rattus said:
Think about it Smart. V and I are constants, therefore Z is a constant. Not so with v(t) and i(t). Does not follow.
Click to expand...
That just leads to the question: How does it not follow?

V and I are constants as is P, but yet we can easily conclude that p(t) is not, and v(t) ? i(t) = p(t). Also note V and I are RMS 'constants' while P is an average 'constant' in the equation...
P = V?I = |V||I|cosθ = volts ? amperes ? power factor​

V, I, P, Z are all mathematical manipulations of the raw data of v(t) and i(t). It stands to reason that if the dot product of Vrms and Irms is an average, namely P, then the complex division of V by I is an average, namely Z. And actually, vector division (follow up) has less complications than does [dot product] vector multiplication (follow up).
 
Smart $

Smart $

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Hameedulla-Ekhlas said:
Ok for something different question something different answer

v(t)/i(t) =(I[Vcos∅ { 1-cos⁡(2wt) }- Vsin∅sin(2wt) ]^2)/(2√(2 ) Psin(wt-∅)sin⁡(wt))

let me clear the question a little bit

b3x7xf.png


Now who can go reverse to prove v(t)/i(t) step by step ?
Click to expand...
Why would we go to the trouble of using your formula when we can simply substitute the formulas of v(t) and i(t)...
v(t) = |v|cos(ωt)
i(t) = |i|cos(ωt+θ)

thus

v(t)/i(t) = |v|cos(ωt) / |i|cos(ωt+θ)​
 
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