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Its set on AC.anorlunda said:I searched this thread. You didn't say whether the multimeter is set to an AC scale or DC scale.
Its set on AC.anorlunda said:I searched this thread. You didn't say whether the multimeter is set to an AC scale or DC scale.
Just the resistor. And what does 0 deg mean here. Zero phase offset from the current but the current is offset from the applied voltage. This is just what you wish to define as t=0erobz said:I must still be confused. If I measure a voltage drop across a resistor, and I know the resistance, can't I calculate the current through resistor? Or is it that the voltage I measure across the resistor is only representative of the portion of current that is in phase with the resistor at 0∘?
Ok, So here is the sticky part. I measured ##24.5 \rm{V}## across the relay, which is NOT just a resistor. It is a series resistor/inductor...I assumed that meant the meter would read the missing ## 0.1 \rm{V}## across the transmission lines if I had directly measured that, but it doesn't sound like that is accurate.hutchphd said:Just the resistor. And what does 0 deg mean here. Zero phase offset from the current but the current is offset from the applied voltage. This is just what you wish to define as t=0
In a series circuit the current is everywhere the same phase (there is no place where it can escape (even a capacitor has no net charge).
Across a resister the voltage is always exactly in phase with the current. Across an isolated inductor/ capacitor the current is 90deg behind/ahead of the voltage.
For a combination the phase varies and depends upon frequency. This gets us to phasors or equivalently (and more usefully) to complex numbers. Worth the effort. Resistance is real, Reactance imaginary, and Impedance thereby complex
Then you are measuring averages. Importantly, a multimeter can never measure the phase information that you need to completely describe what is happening. For example, the ##\phi## in ## I(t)=I_0 \sin(\omega t+\phi )##erobz said:Its set on AC.
With the result above I was just trying to estimate the inductance of the coil, because it wasn't given by the manufacturer to see what I get.hutchphd said:If you are going to estimate something that matters, you usually want to know that the estimate is either reliably an overestimate or an underestimate.
What is it that you really want to know here and which way do you wish to err? For instance the magnitude of the Impedance Z of a resistor R in series with an inductor L is $$|Z| =R \sqrt {1+ \frac {(\omega L)^2} {R^2}} $$ and the magnitude of the steady state current amplitude is $$I_{rms}=\frac {V_{rms}} {|Z|}$$. So ignoring L overestimates the current draw. If L=0.1H then then the over estimate will go like half of ##(\frac {\omega L} R)^2## which is <1%
This is just wrong because j*j =1 (where z* denotes complex conjugate of z). You need to do the work if you want get the correct answers. You had it correct earlier!erobz said:However, when put that way I now notice the reactance will actually decrease with increasing inductance ( since j2=−1), so the current must be larger than 138mA, not smaller.
The sufficient answer is that L is small enough to ignore for considerations of steady state and ignoring increases your safety margin. Yea!erobz said:But I believe its safe to say that worst case the highest current in the circuit is 138mA, since any appreciable inductance will drop the current?
Which post is junk #40?hutchphd said:I went through this because your method of estimating L contains unnecessary errors. If I understand your method. the worst part was trying to find the inductance without using the appropriate form for the impedance. And you compound the problem by taking the difference of two large numbers to obtain a small one. Sorry it is pretty much junk IMHO
This is just wrong because j*j =1 (where z* denotes complex conjugate of z). You need to do the work if you want get the correct answers. You had it correct earlier!
The sufficient answer is that L is small enough to ignore for considerations of steady state and ignoring increases your safety margin. Yea!
I won't declare victory, but I'll retreat. You were annoyed on post #9. I don't want to fry your last nerve! Thanks for the help.hutchphd said:I think so. For the reasons I outlined the approach is really ill-advised for the same reasons that it is unnecessary. If the current is insensitive to L then your method (which essentially measures the current) will not give a robust value for L. Declare victory and retreat (or dive deeper and regroup).
Before I go. What?hutchphd said:This is just wrong because j*j =1
hutchphd said:And you compound the problem by taking the difference of two large numbers to obtain a small one.
hutchphd said:the worst part was trying to find the inductance without using the appropriate form for the impedance