What would happen in this circuit? inductor/reactor series

[Electric-Light]

L1 to L2 = 240v L1 or L2 to N = 120v

L1--[L 318mH]---{Ntrl}---[C 44.2?F]--L2

If my math is right, L1 to neutral is 120ohm therefore 1A +90? and L2 to N is also 1A, but -90?

So, if it was on a MWBC, I would think that the two loads would act independently of each other, but would it short out as soon as you disconnect the neutral as the capacitive and inductive impedance cancel each other out?

 

[gar]

121124-1433 EST

Change the capacitance to 22.1048 mfd and after some transient time, dependent upon all the the initial conditions, and assuming 60 Hz, then a large current will occur. Series resonance.

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[G._S._Ohm]

If I didn't screw up the signs, with equal reactances there will be two amps in the neutral with the neutral in place.

This one is a little tricky.

 

[Electric-Light]

121124-1433 EST

Change the capacitance to 22.1048 mfd and after some transient time, dependent upon all the the initial conditions, and assuming 60 Hz, then a large current will occur. Series resonance.

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Assume 60Hz. I forgot to multiply the pi by 2, so you're right.
So, assume perfect reactor and capacitor each having impedance of 120 ohm at 60Hz. Assume perfectly balanced and frequently is dead on.

Do they behave like separate non power consuming loads with neutral in place, then short out as soon as the neutral is removed?
What would happen if say the neutral has a resistance of 1 ohm?

 

[gar]

121124-1617

I believe Electric-Light wanted series resonance at 60 Hz and his capacitance was double the value for resonance. While the neutral is connected to the common point between the capacitor and inductor and for the reactance values given the neutral current would be 2 A no matter what the frequency was so long as the reactance values were for that frequency.

From my Shure Brothers Reactance Slide Rule for the given inductance and capacitance the approximate resonant frequency is about 42 Hz.

Also note that for the series resonant case that the initial conditions can be specified such that there is zero initial current at switch closure, or a very large current, or anything in between.

.

 

[G._S._Ohm]

What would happen if say the neutral has a resistance of 1 ohm?

Then the current in the neutral would be
120{[1/(1+j120)] - [1/(1-j120)]}
You need to convert to polar coord. and then back to rectangular coord. to reduce this and there are web calculators that will do that.

This is just like being back at school except that I'm married and older and considerably less stressed-out.

I have a feeling you will have more questions. . .

With a center-tapped 24 vac transformer and a large inductor and small capacitor and a milliammeter, you could try a scaled-down version of this.

 

[gar]

121124-1639 EST

The initial conditions at the time of opening the neutral will determine the initial current. Steady state it is large. In the real world you will have some resistance in series with the resonant circuit and this determines the steady state current.

Most of the resistance will be in the inductor. Further there will be saturation or breakdown characteristics that will limit maximum voltage and current.

With 120 ohms of reactance in each phase and 1 ohm resistance in the neutral the neutral point of the capacitor and inductor will deviate from the source neutral by about your 2 A and 1 ohm, 2 V.

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[Electric-Light]

121124-1639 EST

The initial conditions at the time of opening the neutral will determine the initial current. Steady state it is large. In the real world you will have some resistance in series with the resonant circuit and this determines the steady state current.

Most of the resistance will be in the inductor. Further there will be saturation or breakdown characteristics that will limit maximum voltage and current.

With 120 ohms of reactance in each phase and 1 ohm resistance in the neutral the neutral point of the capacitor and inductor will deviate from the source neutral by about your 2 A and 1 ohm, 2 V.

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Well, if the neutral had no resistance, the current through the reactor and capacitor would be infinitely close to what they would be if they were fed from separate windings while with the neutral disconnected, it would be series resonance. What I was asking is what happens if you vary the path of resistance to neutral.

By the way, what would happen if you placed a huge capacitor across the line in a building that matches the secondary impedance of the 75kVA 480 to 208/120 transformer serving it?

Would that same capacitor behave any differently if it was to be placed across a source fed from 208/120 underground feed sourced from multiple MVA class trannies?

 

[gar]

121124-1721 EST

From a qualitative descriptive perspective:

1. When the neutral resistance is small compared to reactive impedances, then you can use the approximation I used above.

2. When the neutral resistance is very large compared to the reactive impedances, then the circuit is very close to the series resonant circuit with a small seires resistance lowering the circuit Q, and thus limiting peak voltage and current.

3. Between these extremes write the loop equations and solve.

By the way, what would happen if you placed a huge capacitor across the line in a building that matches the secondary impedance of the 75kVA 480 to 208/120 transformer serving it?

I believe you are referring to the transformer equivalent series inductance as viewed from the secondary side. You create a series resonant circuit with voltages being limited by the load resistance on the secondary side.

Would that same capacitor behave any differently if it was to be placed across a source fed from 208/120 underground feed sourced from multiple MVA class trannies?

What happens will depend upon what is the equivalent circuit and the resistive load across the capacitor.

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[steve66]

L1 to L2 = 240v L1 or L2 to N = 120v

L1--[L 318mH]---{Ntrl}---[C 44.2?F]--L2

If my math is right, L1 to neutral is 120ohm therefore 1A +90? and L2 to N is also 1A, but -90?

So, if it was on a MWBC, I would think that the two loads would act independently of each other, but would it short out as soon as you disconnect the neutral as the capacitive and inductive impedance cancel each other out?

Its a little suprising, but yes, it would actually short out as soon as you remove the neutral wire.

The neutal wire keeps the inductor and capacitor from trading energy back and forth each cycle (which is what normally happens in a resonant circuit.)

 

[G._S._Ohm]

And the one ohm resistor will have exactly two volts across it (I did this circuit three times).

This is not very intuitive, you have to do the math.

 

[gar]

121126-1516 EST

G._S._Ohm:

Intuitive in this sense.

If you can look at the circuit and see that the neutral current is 2 A with no resistance in the neutral circuit, then for small changes in voltage across a resistor in series with the neutral you still have about 2 A. Thus, 2 A thru 1 ohm is 2 V.

A triangle of 120 x 2 corresponds to an angle of 0.95 degrees. This won't make much change in the 2 A value.

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[G._S._Ohm]

121126-1516 EST

G._S._Ohm:

Intuitive in this sense.

If you can look at the circuit and see that the neutral current is 2 A with no resistance in the neutral circuit, then for small changes in voltage across a resistor in series with the neutral you still have about 2 A. Thus, 2 A thru 1 ohm is 2 V.

A triangle of 120 x 2 corresponds to an angle of 0.95 degrees. This won't make much change in the 2 A value.

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I got angles of about 89.5 degrees.

I think it exactly cancels out. For problems like this I recommend using Excel to carry all these significant figures.

This would have made a good power engineering test question in any case.

 

[Electric-Light]

I got angles of about 89.5 degrees.

I think it exactly cancels out. For problems like this I recommend using Excel to carry all these significant figures.

This would have made a good power engineering test question in any case.

In the real world, I think the capacitor would have to be oversized factoring in for the series reactance of transformer and wiring.

If capacitance was to be gradually added across the line, would it actually suddenly cause a disproportionate and sudden rise in current when it hits a resonance?

Would applying a small AC voltage and calculating the impedance by Vrms/Irms @ assumed 60.0Hz and subtracting the DC resistance provide a good enough data to figure out the reactance? The test would be done with the other side of winding left open.

Would this series inductance value hold true under load? i.e. does the series reactance I figure out above stay true if the transformer is loaded with caps?

 

[G._S._Ohm]

Agreed, yes by definition, yes unless you're working for the NIST.

 

[Electric-Light]

Agreed, yes by definition, yes unless you're working for the NIST.

I wish there was a software sandbox that would simulate very close to what would happen with the real deal.
So an ideal capacity and an ideal inductor canceling each other out on a zero resistance circuit would short out, but what happens as the two approaches ideal? i.e. say capacitive impedance is 100.000ohm at test frequency and inductance is slowly adjusted. Do they act as linear subtraction for the most part, then the current rise exponentially as you reach close to resonance?

 

[steve66]

What would happen if say the neutral has a resistance of 1 ohm?

I'm lagging behind - I just got to the 1 ohm question.

Its really pretty neat - the current through the neutral resistor will stay at exactly 2 amps no matter what resistance you put in the neutral path (assuming its not infinity).

So with a short circuit on the neutral, the current is 2 amps total, with 1 amp through the inductor, and one amp through the capacitor.

As teh neutral resistance is raised, the neutral current stays at 2 amps, but the current through the inductor and capacitor starts rising and just keeps going up as the resistance gets larger.

 

[G._S._Ohm]

I'm lagging behind - I just got to the 1 ohm question.

Its really pretty neat - the current through the neutral resistor will stay at exactly 2 amps no matter what resistance you put in the neutral path (assuming its not infinity).

So with a short circuit on the neutral, the current is 2 amps total, with 1 amp through the inductor, and one amp through the capacitor.

As teh neutral resistance is raised, the neutral current stays at 2 amps, but the current through the inductor and capacitor starts rising and just keeps going up as the resistance gets larger.

At this point I recommend writing the two loop equations for the two loop currents.


Use R for the neutral resistance and 120v, and 120 ohms for both reactances.
Once you've solved for R you can see what everything will do as you vary R from zero to infinity.

120 = I1(j120) + (I1-I2)R
and
120 = (I2-I1)(R) + I2(j(-120))

Using cut and paste might make it easier to keep the sign convention correct.



Very simple formulas can do very weird things.
John Dos Paulos used a simple formula to debunk the idea behind "the Laffer curve."

 

[steve66]

At this point I recommend writing the two loop equations for the two loop currents.
Use R for the neutral resistance and 120v and 120 ohms for both reactances.
Once you've solved for R you can see what everything will do as you vary R from zero to infinity.

Very simple formulas can do very weird things.
John Dos Paulos used a simple formula to debunk the idea behind "the Laffer curve."

That's basically what I've done - except instead of using 2 loop equations, I've used Millmans theorem and plugged it into excel.
Having the neutral current stay at 2 amps while everything else in the circuit changes is pretty weird.

If you still have the circuit set up, maybe you can insert a variable resistor in series with the 1 ohm resistor and verify my result. I'm wondering if I have a math error somewhere.

You would have to be careful you don't overload any components. My calculations show at 20 ohms all the currents are still under 2 amps. If you go any higher with the reisistance, the current increases quickly.

Edit: Using excel, I've tried everything from 0.001 ohms to 100 meg. Even tried adding some L and C in the neutral, but it gets weirder. Most numbers give 2 amps, but some combinations for R & X start changing the neutral current. I wonder what is going on there?

 

[G._S._Ohm]

That's basically what I've done - except instead of using 2 loop equations, I've used Millmans theorem and plugged it into excel.

Having the neutral current stay at 2 amps while everything else in the circuit changes is pretty weird.

If you still have the circuit set up, maybe you can insert a variable resistor in series with the 1 ohm resistor and verify my result. I'm wondering if I have a math error somewhere.

You would have to be careful you don't overload any components. My calculations show at 20 ohms all the currents are still under 2 amps. If you go any higher with the reisistance, the current increases quickly.

I haven't done a breadboard in a long time. If you have a sine wave signal gen. you can use small currents and scale up the result.

There's a hifi set in my basement that gets ratty sound in one channel after a half hour and I still haven't assembled the function gen. chip so I can input a sine into this channel and trace it through.
And there's a car radio/cassette player I have that has waited since 2003 to be installed.


And tonight I will maybe go out with long underwear into 0C weather onto the driveway and finish that head gasket job.
:thumbsup:

You know Pspice is available for free on the Web?