inductive reactance of circuit conductors

[winnie]

1000 feet of awg 1 has calculated inductance of 687000 nH or 0.000687 H see: https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

The inductive reactance is 2 pi f L... or 2 (3.141) (6) (0.000687) = 0.259 ohm http://www.66pacific.com/calculators/inductive-reactance-calculator.aspx

Why is the 0.259 ohm so far from the NEC listed value of 0.0046 ohm?

How does NEC derive the 0.0046ohm? Can you help derive 0.0046 ohm? I've been trying it for 2 hours already.

The key issue here is that circuit conductors are laying side by side with the corresponding return conductor; they are not single conductors in free space.

The magnetic field created by current flow in the conductors thus extends far less into free space.

Since self inductance is all about the magnetic flux created by current flowing in the wire interacting with the wire, when you have less magnetic flux you have lower inductance.

The larger the space between corresponding circuit conductors, the larger the inductance.

-Jon

 

[tersh]

The key issue here is that circuit conductors are laying side by side with the corresponding return conductor; they are not single conductors in free space.

The magnetic field created by current flow in the conductors thus extends far less into free space.

Since self inductance is all about the magnetic flux created by current flowing in the wire interacting with the wire, when you have less magnetic flux you have lower inductance.

The larger the space between corresponding circuit conductors, the larger the inductance.

-Jon

Thanks for this. It made sense many things.

A bit of detour as I need to know something which I've been thinking for more than a week. What kind of magnetic field is produced for bus bar compared to wires? I know that for wires, it's:

But for bus bar where it is part of the shunt of surge protectors, what kind of magnetic field is produced. Remember that the SPD leads must be as short as possible to avoid huge let through voltage seen at the load... see the following image I took from the internet (No, i'm not installing it don't worry):

The breaker where the Siemens First Surge is connected is at bottom. So for loads that is in the top of the panel (say the first breaker below the main breaker). The bus bar would be part of the shunt or leads of the SPD.

Compared it with the breaker the Siemens First Surge is connected is at the top:

Let say the Siemens First Surge uses wires (drawn in red) to connect to the top breaker (I know the best position is to let the Siemens unit be at the top of the panel, but this is just for sake of discussion), then any loads at second breaker at right of the top breaker would use the red wires as shunt or lead part of the SPD.

Bottom line is. Would the red wires or bus bar as part of the SPD leads increase the let through voltage of load at breakers at top of panels?

First I'm fully aware of the following so you don't have to repeat about the nature of SPDs.

For high frequency, the impedance per unit length of the wire is 1 uH/meter.
How about the bus bar, what is the impedance per unit length of it? Higher or lower or the same?

Can bus bar as part of the leads increase or decrease the voltage across the equipment terminals as described above? What do you think?

 

[winnie]

Inductance calculations for things like bus bars tend to be very approximate. I believe that the self inductance of a bus bar will be lower than that of a round wire of the same cross section.

Here is a simple inductance calculator for flat bars:
https://chemandy.com/calculators/flat-wire-inductor-calculator.htm
However I would not trust the results to be very precise.

-Jon

 

[tersh]

Inductance calculations for things like bus bars tend to be very approximate. I believe that the self inductance of a bus bar will be lower than that of a round wire of the same cross section.

Here is a simple inductance calculator for flat bars:
https://chemandy.com/calculators/flat-wire-inductor-calculator.htm
However I would not trust the results to be very precise.

-Jon

I entered:

Enter the Length: 20 cm
Enter the Width: 0.2 cm
Enter the Thickness: 7.35mm

Inductance: 170.66nH

Compare it to a round wire https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

Diameter of Wire: 7.35mm
Length of Wire: 20cm

Inductance: 158nH

The inductance of round wire of same cross section is smaller.

What example do you have that gave you the inductance of rectangular bar as smaller?

The magnetic field or flux for round wire is:

Do you have illustrations of the magnetic field/fluxes of bus bars:

Is it around the whole block itself?

 

[tersh]

Inductance calculations for things like bus bars tend to be very approximate. I believe that the self inductance of a bus bar will be lower than that of a round wire of the same cross section.

Here is a simple inductance calculator for flat bars:
https://chemandy.com/calculators/flat-wire-inductor-calculator.htm
However I would not trust the results to be very precise.

-Jon

The values are close.

Whatever. Do you agree the following factory SPD placement inside the panel is a bad design?

By putting the SPD below in the panel. The leads would be very long (the length of the bus bar), so for breakers above the panel (shown in green). It would result in much greater voltage across the loads or equipment terminals (as detailed below). Right? I wonder if Siemens designed it that way, what were they thinking? Unless it is possible the bus bar cancel the magnetic field or not part of the leads of the SPD, how?

 

[winnie]

It is very easy to make perfect the enemy of the good.

All other things being equal, I agree that putting the SPD 'ahead' of the branch circuit breakers is better than putting the SPD at the 'end' of the bus.

However I would hesitate to call this a bad design, since the inductance across the length of the bus bars is going to be far lower than the inductance and resistance of the branch circuit conductors.

One post above, when you compared the inductance of two different conductors, you used conductors with different cross section. You might try comparing say a 9cmx1cm bar with a 3cmx3cm bar with a round conductor of the same cross section. But as I said, those calculators are only approximations.

-Jon

 

[tersh]

It is very easy to make perfect the enemy of the good.

All other things being equal, I agree that putting the SPD 'ahead' of the branch circuit breakers is better than putting the SPD at the 'end' of the bus.

However I would hesitate to call this a bad design, since the inductance across the length of the bus bars is going to be far lower than the inductance and resistance of the branch circuit conductors.

One post above, when you compared the inductance of two different conductors, you used conductors with different cross section. You might try comparing say a 9cmx1cm bar with a 3cmx3cm bar with a round conductor of the same cross section. But as I said, those calculators are only approximations.

-Jon

Ah you mean cross section as in the area

Area = pi r^2
given the diameter of the round wire is 7.35mm, then
Area = 3.141 x (7.35/2)^2=42.43 sq.mm

To get rectangular cross section equal to it. Let's just get square root of it to get 6.5mm
https://chemandy.com/calculators/flat-wire-inductor-calculator.htm

Enter the Length: 20 cm
Enter the Width: 0.65 cm
Enter the Thickness: 6.5 mm

Inductance: 157.64nH

compare this to inductance of round wire of same cross section (area):
https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

Diameter of Wire: 7.35mm
Length of Wire: 20cm

Inductance: 158nH

Almost the same although the rectangular one has slightly smaller inductance as you said.

Now if you will compare it to the inductance of the actual bus bar which has larger width of say 25mm

Enter the Length: 20 cm
Enter the Width: 2.5 cm
Enter the Thickness: 6.5 mm

Inductance: 123nH​

So a larger actual bus bar cross section (compared to the actual round wire (awg 10) used in the Siemens SPD lead of smaller cross section) have lower inductance.

I wrote all of the above so I can refer to it in the future as I've been pondering it for a month.

Also I noticed that in the wire and rectangular bar calculator, if you make the cross section larger, the inductance is lower.

What principle caused larger cross sections to have smaller inductance? Any idea? Thanks.



​

 

[GoldDigger]

Ah you mean cross section as in the area

Area = pi r^2
given the diameter of the round wire is 7.35mm, then
Area = 3.141 x (7.35/2)^2=42.43 sq.mm

To get rectangular cross section equal to it. Let's just get square root of it to get 6.5mm
https://chemandy.com/calculators/flat-wire-inductor-calculator.htm

Enter the Length: 20 cm
Enter the Width: 0.65 cm
Enter the Thickness: 6.5 mm

Inductance: 157.64nH

compare this to inductance of round wire of same cross section (area):
https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

Diameter of Wire: 7.35mm
Length of Wire: 20cm

Inductance: 158nH

Almost the same although the rectangular one has slightly smaller inductance as you said.

Now if you will compare it to the inductance of the actual bus bar which has larger width of say 25mm

Enter the Length: 20 cm
Enter the Width: 2.5 cm
Enter the Thickness: 6.5 mm

Inductance: 123nH​

So a larger actual bus bar cross section (compared to the actual round wire (awg 10) used in the Siemens SPD lead of smaller cross section) have lower inductance.

I wrote all of the above so I can refer to it in the future as I've been pondering it for a month.

Also I noticed that in the wire and rectangular bar calculator, if you make the cross section larger, the inductance is lower.

What principle caused larger cross sections to have smaller inductance? Any idea? Thanks.



​

Basically when the current is spread out across a large area, the self-inductance will be lower because there will not be a concentrated (high strength) magnetic field close to the centerline of the current flow. The region very close to the wire contributes correspondingly more to the magnetic induction that the distant field contribution.

 

[tersh]

Basically when the current is spread out across a large area, the self-inductance will be lower because there will not be a concentrated (high strength) magnetic field close to the centerline of the current flow. The region very close to the wire contributes correspondingly more to the magnetic induction that the distant field contribution.

I see.

By the way. The bus bar has much more current moving in it, although just 60 Hz.

But when there is a high frequency surge and the bus bar is part of the shunting leads. Won't it contribute to huge inductance effect (say the bus bar has 100 ampere of 60 Hz current cycling in it)? Would this be seen by the SPD as like having huge inductance?

 

[GoldDigger]

I see.

By the way. The bus bar has much more current moving in it, although just 60 Hz.

But when there is a high frequency surge and the bus bar is part of the shunting leads. Won't it contribute to huge inductance effect (say the bus bar has 100 ampere of 60 Hz current cycling in it)? Would this be seen by the SPD as like having huge inductance?

The amount of 60Hz current in the bus bar at any given time has no effect on the inductance seen by a high frequency applied current.
The only thing that would cause the inductance itself to be different would be any change in current distribution caused by skin effect. The inductance would be the same, but the inductive reactance, being proportional to frequency, would be much higher for the higher frequencies.
The inductive reactance of the bus bar in the SPD current path might be significant, but it would be smaller than the reactance of the same length of cylindrical wire. Shorter is better, but if it has to be longer, bus bar is better than wire.

 

[winnie]

Enter the Length: 20 cm
Enter the Width: 0.65 cm
Enter the Thickness: 6.5 mm

Inductance: 157.64nH

compare this to inductance of round wire of same cross section (area):
https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

Diameter of Wire: 7.35mm
Length of Wire: 20cm

Inductance: 158nH

Almost the same although the rectangular one has slightly smaller inductance as you said.

Yes, but in the above you are comparing a _square_ conductor to a round conductor of similar area, using two different equations. The round conductor equation might be exact from theory, but my guess is the rectangular conductor equation is an approximation.

Now using the _same_ cross section, look at conductor:
20cm long
6.5cm wide
0.65mm thick

same cross section, but the calculated inductance drops to 95nH.

-Jon

 

[tersh]

It is very easy to make perfect the enemy of the good.

All other things being equal, I agree that putting the SPD 'ahead' of the branch circuit breakers is better than putting the SPD at the 'end' of the bus.

However I would hesitate to call this a bad design, since the inductance across the length of the bus bars is going to be far lower than the inductance and resistance of the branch circuit conductors.

One post above, when you compared the inductance of two different conductors, you used conductors with different cross section. You might try comparing say a 9cmx1cm bar with a 3cmx3cm bar with a round conductor of the same cross section. But as I said, those calculators are only approximations.

-Jon

There is something Ive been pondering for over a month so hope you can clarify this once and for all. Supposed there was 40 meters of awg 1 wires from the service entrance to the townhouse panel inside the 5 townhouse compound and you couldnt put the SPD at the service entrance but only at the main panel inside the townhouse after 40 meters of wire. Would the 40 meters wires increase the voltage at the loads or equipment terminals after the SPD or would the 40 meters of wire act as additional source impedance that can lower the residual voltage at the loads terminal?

Last year you told me and i understood the essence that..

"If you are protecting from 'excessive voltage between your supply conductors', then you need*impedance*in your supply circuit 'upstream' of the SPD. The SPD acts as a voltage divider with that supply*impedance*to reduce the peak voltage at the terminals of the SPD. Any 'downstream'*impedance*between the SPD and the load will act to reduce the voltage further, so for this particular type of surge*impedance*between the SPD and the load is not an important factor."

 

[winnie]

There is something Ive been pondering for over a month so hope you can clarify this once and for all. Supposed there was 40 meters of awg 1 wires from the service entrance to the townhouse panel inside the 5 townhouse compound and you couldnt put the SPD at the service entrance but only at the main panel inside the townhouse after 40 meters of wire. Would the 40 meters wires increase the voltage at the loads or equipment terminals after the SPD or would the 40 meters of wire act as additional source impedance that can lower the residual voltage at the loads terminal?

I would expect the 40m of conductor to act as an additional source impedance reducing voltage at the load terminals.

I could imagine a situation where current flowing in this 40m of conductor gets interrupted at the load end, causing an inductive kick and high voltage at the load terminals, but then you are looking at some sort of switching event in your equipment causing voltage spikes that affect other equipment.

-Jon

 

[gar]

190324-0919 EDT

Ponder this --- a transmission line 1/4 wavelength long with a short at one end reflects that low impedance to a high impedance at the open end. 300 meters is one wavelength at 1 MHz, or 1/4 wavelength occurs at 2.5 MHz for 30 meters.

A current at the shorted end produces a high voltage at the open end.

A transmission line sourced and terminated by its characteristic impedance does not have this multiplication.

A shorted 1/4 wavelength transmission line makes a nice resonator for an oscillator.

.

 

[tersh]

I would expect the 40m of conductor to act as an additional source impedance reducing voltage at the load terminals.

I could imagine a situation where current flowing in this 40m of conductor gets interrupted at the load end, causing an inductive kick and high voltage at the load terminals, but then you are looking at some sort of switching event in your equipment causing voltage spikes that affect other equipment.

-Jon

Ok thanks Winnie.

This whole week. I'm trying to figure something out and it consumed half of my nights. Maybe you know the answer, so hope you can help. This will be my last and final inquiry about it all.

Are you familiar with the f and M Point and Point Calculation Procedure to get the bolted short circuit current?

http://www.cooperindustries.com/con...rary/BUS_Ele_Tech_Lib_Electrical_Formulas.pdf

The problem is the result doesn't match the method where you added the transformer and conductor impedances.

First I'm aware how the transformer impedances are added to the wiring impedances. Someone emphasized to me that:

"Transformer resistive and reactive components

The transformer nameplate states the impedance as a percentage. Knowing only the percentage does not allow the resistive and reactive components, R and X, to be determined. Therefore the result of the above calculation is the combined resistance and reactance, Z. Often nameplates state X/R as well as percentage. That allows Z to be calculated when necessary. Another alternative is to find a listing of typical X/R for various transformer sizes. At this kVA level, X/R is likely more than 5 and therefore the impedance can be considered to be mostly inductive reactance.

If any added external impedance is mostly inductive, it can simply be added. If it is mostly resistive, it can be added using the square root of the sum of squares. In the case of adding the impedance of half of an additional winding, X/R is assumed to be the same for both components and can be simply added."​

(end quote)

So to add the inductive reactance of transformers to the wiring conductors. You need to use the formula:​

Using direct impedance computations from the table below. The bolted short circuit current varies the ones with f and M (point to point method) by a wide margin (shown below). ​

I still don't know why and arc flash experts don't know how the f and M are derived.

Let's compute:​

https://peritoselectricos.mx/wp-con...ay-Electric-Power-Systems-in-Commercial-B.pdf

Table 65 for AWG 1 at 1000 feet for Nonmetallic conductors and Several 1/C Nonmagc conductor Data​

Resistance is 0.129 ohm
Reactance is 0.0342

For 1 foot:
resistance is 0.129 ohm * 1/1000= 0.000129 ohm
reactance is 0.0342 * 1/1000 = 0.0000342

Impedance = sqrt (0.000129^2+0.0000342^2) = 0.0001335 ohm

1/Z = 1/0.0001335 = 7493 which is the C listed for AWG 1 Cooper nonmetallic 600v in the cooper pdf (see the first url above)

now going to the formula f = [1.732 * L * I] /[ N * C * E(l-l)]

entering the above data,
f is 0.347,
M = 0.742
Isc (bolted scc) = 11597 amp

This spreadsheet produced the results too: https://electrical-engineering-port...electrical-software/fault-current-calculation

Now if you will compute it directly by getting the impedances of the wires and transformers and dividing 240v by the total impedances to get the bolted short circuit current, the result doesn't match. ​

Here is the computation of the direct impedances addition and getting the current by dividing the voltage by the impedance.

From the same impedance table above where the C=1/Z came from, and adding the transformer impedance to the wire inductive reactance side of it to get overall bolted impedance:

For 20 feet AWG 1:
resistance is 0.129 ohm * 20/1000= 0.000258 ohm
reactance is 0.0342 * 20/1000 = 0.0000684
Impedance = sqrt (0.000258^2+(0.0000684+0.01536)^2) = 0.01625 ohm
Transformer impedance of 0.01536 comes from 240^2/75000 * 0.02

240v/0.01625 = 14769 amp.

This differs from the IEEE f and M point to point formulas by wide margin (14769A vs 11597a).

I assume the X/R of the transformers is purely inductive. Even if you divided it among the resistance and inductive reactance. It won't lower the bolted short circuit current unless you assume X/R is purely resistive (which isn't true because the transformer impedance is most inductive reactance).

Since this addition of impedances and getting the bolted short circuit current by dividing the voltage by the total bolted impedance is more direct and accurate. Then is the f and M not accurate? How is f and M derived?

Remember to add the inductive reactance of transformers to the wiring conductors. You need to use the formula:

The inductive reactance must be added inside the X component. Now look at the IEEE bolted short formulas of f and M

The C is just 1/Z of the wiring conductor only. You are supposed to add the transformers inductive reactance inside the X component of C. What the formula f does is adding it outside.

Why did they add it outside? It makes the bolted short circuit current after 20 feet of wiring conductors much lower than in reality.

What do you think?​

 

[tersh]

Ok thanks Winnie.

This whole week. I'm trying to figure something out and it consumed half of my nights. Maybe you know the answer, so hope you can help. This will be my last and final inquiry about it all.

Are you familiar with the f and M Point and Point Calculation Procedure to get the bolted short circuit current?

http://www.cooperindustries.com/con...rary/BUS_Ele_Tech_Lib_Electrical_Formulas.pdf

The problem is the result doesn't match the method where you added the transformer and conductor impedances.

First I'm aware how the transformer impedances are added to the wiring impedances. Someone emphasized to me that:

"Transformer resistive and reactive components

The transformer nameplate states the impedance as a percentage. Knowing only the percentage does not allow the resistive and reactive components, R and X, to be determined. Therefore the result of the above calculation is the combined resistance and reactance, Z. Often nameplates state X/R as well as percentage. That allows Z to be calculated when necessary. Another alternative is to find a listing of typical X/R for various transformer sizes. At this kVA level, X/R is likely more than 5 and therefore the impedance can be considered to be mostly inductive reactance.

If any added external impedance is mostly inductive, it can simply be added. If it is mostly resistive, it can be added using the square root of the sum of squares. In the case of adding the impedance of half of an additional winding, X/R is assumed to be the same for both components and can be simply added."​

(end quote)

Btw.. the above quote came from an EE at physics stack exchange who was not familiar with the f and m point to point calculation method. Is this not taught to all EE?

He was just teaching me how to get the shorted circuit current of two open delta transformers. He never get used to adding the impedances of transformers and circuit conductors (wiring) so for any other EE here familiar with them, kindly extend an analyzing hand because of the possibility the m and f method may not be entirely correct.

About how the short circuit current of the transformers is calculated. The following are the details (just for background). I just wanted to get the bolted short circuit current after 20 feet of awg 1 wire. How would you do it on your own? Winnie, Gar, any veteran EE? Just use the 240/120v single phase data in my last message for simplicity to determine why the f and c method results vary greatly from the direct method of getting all impedances (transformers and wiring) and getting the bolted current from it (by dividing the voltage with the total Z).

single transformers data:

The EE in PSE who is not familiar with the point to point fault current method just shared how the transformer short circuit current is determined:

"

To calculate the current in a short circuit between point B and the center tap of the A to C winding I would first calculate the winding impedance. The full 240 volt winding is rated 75 kVA. The impedance of a load that would draw 75 kVA is 240^2/75000 = 0.768 ohms. The transformer impedance is 2.59% or 0.0259 x 0.768 = 0.0199 ohms. The total impedance of a 120 V winding in series with a 240 V winding is 1.5 x 0.0199 = 0.0298 ohms. The short circuit current would then be 208/.0298 = 6971 amps.
I used the percent impedance to calculate impedance in ohms because the percent impedance is based on 75 kVA at 240 volts and the combination of windings results in 208 volts.
It seems easier and more clear to used a fixed ohmic value when dealing with one single-phase transformer and part of another connected to three-phase power.I believe the equivalent circuit is a shown below. Each transformer or part of a transformer is represented as an ideal voltage source and an impedance. The impedance of the 3-phase source is assumed to be zero."

(His next paragraphs was my quotes in my last message)

That's only as far as the EE knew. Again he didn't know the f and M formulas. So I'm stuck for two weeks pondering on all of these.

Who knows them? Do you Gar, Winnie? Why is the result of the bolted short circuit current solved by that method vary widely from the natural formulas of just getting transformers and wiring conductors and dividing voltage from the impedance to get the bolted short circuit current?
​

 

[tersh]

The point to point fault calculation method is also shared by Mike Holt himself in:

https://www.mikeholt.com/freestuff-shortcircuitcalculations.php

How is f and m derived?

The result varies even to another NEC method mentioned in :

https://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/

Here's it's more intuitive.

How can two results vary by over 30%?

 

[tersh]

I am finally able to derive f and M and solve it by using different formula (direct method) but same results after trying different things in excel.

I adjusted the same formula I used before and it matched.

For 2 pairs of 20 feet AWG 1.. the conductors must be 40 feet instead of 20 feet, and using the same resistance and reactance from the IEEE table (where they got the C= 1/Z).

resistance is 0.129 ohm * 40/1000= 0.00516 ohm
reactance is 0.0342 * 40/1000 = 0.001368 ohm

Z = sqrt (0.00516^2 + 0.001368^2) = 0.005338 ohm

transformer impedance + conductor impedance = 0.01536 + 0.005338 = 0.020698 ohm

240/0.020698 ohm = 11595A bolted short circuit current... this result is identical to the one using M and f (shown in excel below)

(note Transformer impedance of 0.01536 comes from 240^2/75000 * 0.02)

So what they did is to add by brute force the impedance of the transformer and conductor...

Adding Z1 + Z2 is based on the assumption that X/R is the same for both. Is it?

If the X/R ratio is not equal, then the equation is not the one for it.
What do you think? Can you just add Z1 + Z2 if the X/R ratio is not the same for both?

I'm reviewing vector algebra and it doesn't seem to hold. Correct me where I comprehended wrong.

Identical results to:

 

[tersh]

Anyway. Here is how exactly the formula of f and M are derived. I can't believe I missed it for a week. That's the problem when one didn't pay attention to math in college

After looking at many entries in excel and trying different data. I finally saw the pattern. It's simply:​

They are simply two impedances where the Z of wire is divided to the Z of transformers. Why is shown at bottom. But first, remember C = 1/Z, then it's in 1 foot.. so you multiply it by the Length to get the total wiring impedance, now if there is say 2 conductors per phase, then it's divided by 2 simply because for parallel resistors, you get 1/2R for two resistors in parallel.

Now for the transformer impedance. It's simply R=V/I or Z = E(l-l) / I(l-l)

The reason the transformer impedance is below with wiring impedance above in the divide sign is the following.

The formula is just telling you that if you have a certain impedance at the short circuit current, Isc. Then if you add the wiring impedance to the transformer impedance, then the bolted short circuit current would be inversely proportional to the total over the partial Isc. Inversely proportional because the current goes down if impedances are high. Using the same data in last message, then it's simply

Zwire = Z * 2L/ n = 0.000133458 * 2 (20)/1
= 0.0053383ohm

Ztransformer = 240^2/75000 * 0.02 =

0.01536 ohm​

​

What the M = 1/(1+f) formula does is simply getting the inverse proportional algebra...

That is

I(trans)/Ibolted= (0.01536 + 0.0053383) / 0.01536

given I(trans)sc of 15625A,

then solving for I (bolted scc) would give you 11595A. Exactly the same as the excel sheet in last message because this is all there is to it.

Adding Z1 + Z2 is based on the assumption that X/R is the same for both transformer and wiring. If X/R is not the same, then the bolted short circuit currents computed would be incorrect.

So my only questions now are. How often are X/R ratio of transformers and wiring the same? If seldom, then what formulas must you use instead?

Hope when mentioning the formulas of f and M, they can put disclaimer that it's only if the X/R of both transformers and wiring is the same. A veteran EE emphasized to me that's the only time you can add Z1 + Z2. Otherwise you need to use Z =sqrt(R^2 + X^2) where the components must be separately added.
​