Coil Voltage/Current: Understand Resistance in Relay Circuits

[erobz]

Sorry where did the 26.5V come from? And why are you adding it in phase ?

The 26.5 volts was a goof - me jumbling numbers up. The source voltage is 24.6 V.

And why are you adding it in phase ?

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 $0^{\circ}$?

I'm probably being too loose with my analysis.

What I'm asking: Is the voltage I'm measuring on the multimeter actually $V_r = |I |r \cos \theta$, where $\theta$ is the phase angle?

 

[erobz]

I'm guessing Kirchoff's is actually this:

$$ 0 = V_s \angle 0^{\circ} - I \angle \theta^{\circ} r \angle 0^{\circ} - I \angle{ \theta^{\circ} } R \angle 0^{\circ} - I \angle \theta^{\circ} \omega L \angle 90^{\circ} $$

This AC electricity stuff is tricky business. My hat is off to you guys!

 

[anorlunda]

This AC electricity stuff is tricky business. My hat is off to you guys!

If you could learn to work with complex numbers, you will find that the equations are exactly the same as the DC case. Of course, scalar is simpler than complex [no pun intended], but those trigonometric forms you showed are much more difficult than complex.

 

[erobz]

If you could learn to work with complex numbers, you will find that the equations are exactly the same as the DC case. Of course, scalar is simpler than complex [no pun intended], but those trigonometric forms you showed are much more difficult than complex.

I think my main problem here is I don't know what I'm actually measuring with my multimeter. ( I'm rusty on complex number tricks - but conceptually they aren't a problem).

 

[anorlunda]

hink my main problem here is I don't know what I'm actually measuring with my multimeter.

I searched this thread. You didn't say whether the multimeter is set to an AC scale or DC scale.

 

[erobz]

I searched this thread. You didn't say whether the multimeter is set to an AC scale or DC scale.

Its set on AC.

 

[hutchphd]

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∘?

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

 

[erobz]

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

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.

 

[anorlunda]

Its set on AC.

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 $I(t)=I_0 \sin(\omega t+\phi )$

To entirely see what is happening, you need an oscilloscope, not a multimeter.

 

[erobz]

Ok, How do you feel about this:

Since we suspect $L$ to be small the voltage I measured across the relay $V_{relay}= 24.5 \rm{V}$ is approximately in phase with the source voltage $V_s = 24.6 \rm{V}$ . It follows that:$$V_{relay} = \frac{R+ \omega L j}{ R + r + \omega L j}\cdot V_s $$

Multiply the denominator by its complex conjugate divide by $V_s$:

$$ \frac{ V_{relay}}{V_s} = \frac{ R ( R+r) - R \omega L j + (R+r) \omega L j + ( \omega L )^2 }{ ( R+r)^2 + ( \omega L )^2} $$

Since $R \gg r$ the complex terms in the numerator approximately cancel leaving:

$$ \frac{ V_{relay}}{V_s} = \frac{ R ( R+r)+ ( \omega L )^2 }{ ( R+r)^2 + ( \omega L )^2} $$

with
$f = 60 \rm{Hz}$ or $\omega = 377 \rm{ \frac{1}{s}}$
$R = 177 \Omega$
$r = 0.8 \Omega$
$V_s = 24.6 \rm{V}$
$V_{relay} = 24.5 \rm{V}$

I get that $L \approx 0.17 \rm{H}$

I feel like I mostly followed the rules there, with some (seemingly to me) reasonable assumptions? Either way, I'll let it be. That was my best go at it with the tools I have.

 

[hutchphd]

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 and R=177ohms then the over estimate will go like half of $(\frac {\omega L} R)^2$ which is <1%

 

[erobz]

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%

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.

I was initially concerned about "hidden" current draw because of the inductor. As you point out, just looking at the impedance of an LR circuit should alleviate those concerns. Also the earlier ODE already removed those fears.

I was trying just to learn more by attempting to calculate the inductance. Trying to figure out whether it's an over/underestimate would be pretty involved since I made multiple simplifying assumptions?

Perhaps as a scientist that is an issue, but that's got to be par for the course as a low-grade MET playing with electricity.

 

[hutchphd]

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

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.

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!

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?

The sufficient answer is that L is small enough to ignore for considerations of steady state and ignoring increases your safety margin. Yea!

 

[erobz]

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!

Which post is junk #40?

 

[hutchphd]

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).

 

[erobz]

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).

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.

 

[erobz]

This is just wrong because j*j =1

Before I go. What?

$j^2 = -1$ Its right in the textbook I'm looking at titled "Introductory Circuit Analysis".

$j = \sqrt{-1}$ . They don't use $i$ for obvious reasons.

 

[erobz]

And you compound the problem by taking the difference of two large numbers to obtain a small one.

You mean you don't like $( R+r)\omega L - R \omega L < \omega L \approx 0$ for this circuit?

For $L=0.1$ the term preceding it on the left is 3 orders of magnitude larger, and the one on the right of it is nearly 2 orders of magnitude larger than it. I don't see why adding $37$ to $33,000$ matters?

 

[erobz]

the worst part was trying to find the inductance without using the appropriate form for the impedance

I "dug deeper" and realized what you meant here... Why do smart people have to be so cryptic! You could have pointed out that $Z_{relay} \neq R + \omega L j$. I would have realized the goof in a second.

This time I used the proper polar form:

$$ \mathbf{V_{relay}} = \frac{ \mathbf{Z_{relay}} }{ \mathbf{Z_T }} \mathbf{V_s} $$

$$ \begin{align} \mathbf{ V_{relay} } &= \frac{ \sqrt{ R^2 + ( \omega L )^2} \angle \theta_{relay} \cdot V_s \angle 0^{ \circ } } { \sqrt{(R+r)^2 + ( \omega L)^2} \angle \theta_T } \tag*{} \\ \quad \tag{1} \\ &= V_s \sqrt{ \frac{ R^2 + ( \omega L )^2 }{ (R+r)^2 + ( \omega L )^2 } } \quad \angle \left( \theta_{relay} -\theta_T \right) \tag*{} \end{align} $$

Where:

$\theta_{relay} = \arctan \left( \frac{ \omega L}{R} \right)$
$\theta_{T} = \arctan \left( \frac{ \omega L}{R+r} \right)$

Since $R \gg r \implies \theta_{relay} -\theta_T \approx 0$, we can just work with the magnitudes ( Edit: in hindsight it appears the phase angle is irrelevant to finding the magnitude of $L$ - it could very well be large - ignoring it here is not an approximation on $L$).

$$ V_{relay} = V_s \sqrt{ \frac{ R^2 + ( \omega L )^2}{ (R+r)^2 + ( \omega L)^2} } $$

Solve that for $L$ with all the values above and $L \approx 0.15 \rm{H}$

In the end, what I did on the first attempt didn't turn out to be all that bad of an approximation, (assuming I've got this one correct), but none the less wrong.

EDIT: What I did in post #40 was not wrong. The impedance in complex form is $Z_{relay} = R + X_L j$. When @hutchphd said I used the "wrong form" of impedance, I assumed they meant I should be using the polar form where $\mathbf{Z} = |Z| \angle \theta$. As it stands, as an approximation for this circuit where $R \gg r$, $r<1$, and $L$ is not large I think that is evident from the results that #40 was not "a junk approximation" after all.

That being said, this result is more robust.