…the inductance of the CMC is responsible for the CM
attenuation.
and…
A practical choke is RESISTIVE, not INDUCTIVE.
Emphatic statements indeed.
They are very unlikely to both be correct, and it is possible neither applies generally.
This article calculates the common mode current Icm for three Zcm scenarios of a current balun or common mode choke applied to my own antenna system documented at Equivalent circuit of an antenna system at 3.6 and 7.1MHz. The three choke Zcm values are:
The measured antenna system common mode impedances Zc are:
3.6MHz: 59+j167Ω; and
7.1MHz: 134j44Ω.
Be aware that most backoftheenvelope ohms law analyses are seriously flawed, but in this case we have the common mode impedance of the antenna system measured at the grounded shack patch panel
F  Zc \ Zcm  1500+j1500Ω  1500+j0Ω  0+j1500Ω 
3.6MHz  59+j167Ω  438µA  638µA  600µA 
7.1MHz  134j44Ω  457µA  612µA  684µA 
Above a summary of the magnitude of common mode currents Icm for 1V of common mode drive at the patch panel.
Keep in mind that the common mode current is usually a standing wave in a three dimensional context and cannot be adequately described by a current at a single point.
The broad conclusions for this scenario are:
There is no evidence in the analysis of this real antenna scenario that choke resistance or choke reactance dominates Icm, neither of the online experts are correct in this case.
Firstly, loss means PowerIn/PowerOut, and can be expressed in dB as 10log(PowerIn/PowerOut). For a passive network, loss is always greater than unity or +ve in dB.
\(loss=\frac{PowerIn}{PowerOut}\\\)Some might also refer to this as Transmission Loss to avoid doubt, but it is the fundamental meaning of loss which might be further qualified.
So, lets find the two quantities in the right hand side using ‘powerwaves’ as used in S parameter measurement.
s11 and s21 are complex quantities, both relative to port 1 forward power, so we can use them to calculate relative PowerIn and relative PowerOut, and from that PowerIn/PowerOut.
PowerIn is port 1 forward power less the reflected power at port 1, \(PowerIn=P_{fwd} \cdot (1s11^2)\).
PowerOut is port 2 forward power times less the reflected power at the load (which we take to be zero as under this test it is a good 50Ω termination), \(PowerOut=P_{fwd} \cdot s21^2 \).
So, we can calculate \(loss=\frac{PowerIn}{PowerOut}=\frac{\frac{PowerIn}{P_{fwd}}}{ \frac{PowerOut}{P_{fwd}}}=\frac{1s11^2}{s21^2}\)
Noelec makes a small transformer, the Balun One Nine, pictured above and they offer a set of s11 and s12 curves in a back to back test. (Note: back to back tests are not a very reliable test.)
Let’s work an example.
From the chart marker at 0.2MHz, s11=2.59dB and S21=6.00dB, enter the data into Calculate Loss from s11 and s21.
Above, the calculation. The magnitude of each of s11 and s21 is required by the calculator, the angle of each is not used.
One can approximate the loss of a single transformer by dividing this loss equally between the two cascaded transformers, giving a loss of 1.3dB per transformer.
The MismatchLoss is also calculated. In this case, s21 (6dB) has two main components, MismatchLoss (3.476dB) and Loss (2.524dB). If you were to hook this up between a 50Ω signal generator and 50Ω power meter, inserting the back to back transformer power would reduce the power meter reading 6dB (over a direct connection, partly due to MismatchLoss and partly due to Loss (conversion of RF to heat in the core and windings).
]]>One method is to measure the input impedances of a section of line with both a short circuit and open circuit termination. From Zsc and Zoc we can calculate the Zo, and the complex propagation constant \(\gamma=\alpha + \jmath \beta\).
Calculation of Zo is quite straightforward.
The solution for γ involves the log of a complex number \(r \angle \theta\) which is one of the many possible values \(ln(r) + j \left(\theta + 2 \pi k \right)\) for +ve integer k. Conveniently, the real part α is simply \(ln(r) \). The real part of γ is the attenuation in Np/m which can be scaled to dB/m, and the imaginary part is the phase velocity in ^{c}/m. The challenge is finding k.
So, let’s measure a sample of 14×0.14, 0.22mm^2, 0.5mm dia PVC insulated small speaker twin.
Above is the nanoVNA setup for measurement.
The transformer seen above is a 1:1 transformer as described at A 1:1 RF transformer for measurements – based on noelec 1:9 balun assembly. The system was OSL calibrated at the pin socket underneath the grey connector.
Scans were made from 160MHz of the 1m section of line with short circuit and open circuit termination.
Above is the short circuit scan, the local dip around 30MHz would appear to be an artifact of the test environment, so that area will be avoided.
Above is the open circuit scan which looks generally smooth.
Both scans were saved as .s1p files.
Calculations will be offered at 40 and 14MHz.
Above, the results. Zo is 142j3.5Ω, and matched line loss MLL is 0.3dB/m. This MLL is quite a deal higher than you might expect, even of PVC insulated cables.
Because the line is well less than \(\frac{vf \: \lambda}{4}\) (guessing at the likely value of vf), we can calculate vf simply from the imaginary part of γ for k=0.
PVC / copper speaker twin 0.22mm^2 @ 40.53MHz 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.000500 m 
Spacing  0.001700 m 
Velocity factor  0.660 
Loss tangent  5.000e2 
Frequency  40.530 MHz 
Twist rate  0 t/m 
Length  1.000 m 
Results  
Zo  150.75+j2.90 Ω 
Velocity Factor  0.6600 
Length  73.742 °, 0.204839 λ, 1.000000 m, 5.054e+3 ps 
Line Loss (matched)  0.346 dB 
Above, a TWLLC model was constructed to have about the same MLL. The model would suggest that dielectric loss is quite high. In my experience, PVC LossFactor ranges widely and would seem to depend on pigments, plasticisers and fillers used in manufacture.
Above is a calc from the same sweep, but at 14MHz. MLL is pretty appalling for this sample, yet I see people recommending this stuff and stuff like it for feed line for a dipole, particularly for portable work. Larger size PVC twin lines are better, but never good. The figure8 cable / zip cord dipole of ham lore is one of those things that doesn’t pass analysis.
]]>The behavior of a Guanella 1:1 balun can conveniently be separated into its concurrent common and differential modes.
It is the differential mode that is of most interest when it comes to conductors.
Let’s recall the meaning of InsertionVSWR:
Insertion VSWR is the VSWR looking into the balun with a matched load (termination) on its output, it is a measure of imperfection of the balun.
If the application is one where the transmission line has low standing wave ratio, then you might not want the transmission line embedded in the balun to not degrade that, and the simplest solution is to use the same Zo line within the balun.
Most comonly, it is low InsertionVSWR wrt 50Ω that is the target. The simplest way to achieve 50Ω line in the balun is to use commercial coax. Analysing transmission line loss is relatively simple, use a good line loss calculator.
Ordinary solid polyethylene dielectric coax works fine up to 60°, and with good braid cover even seems to survive being bent to half the specification radius.
PTFE coax has higher temperature withstand, but is typically much stiffer.
Another option is Cujack.
I would avoid ANY foamed dielectric cable, ANY foil shield cables, ANY copper clad steel centre conductors.
Though lots of authors show ways to create nominal 50Ω lines with two or four wires, they are rarely close to 50Ω or uniform, and they are likely to not have the voltage withstand of most practical coaxes. See On use of enameled wire in transmitting baluns for more discussion.
A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail #2 discusses an example where even quite small pigtails on the coax produce measurable InsertionVSWR. For most transmitting purposes, the small defect is not an issue, but may be more important in a measurement system.
For other applications, inherent impedance transformation it a low priority issue, often because it can be offset somewhere else, like in an ATU. When such a balun is used with an ATU, is will often be the case that on the transmission line within the balun for different loads:
Damage may occur due to overheating, core loss and wire loss are the ‘long term’ contributors. Damage may also result from insulation breakdown. Some insulation breakdowns do not cause permanent damage and some do by creating a carbon track in not much more than an instant which degrades the breakdown voltage, often substantially.
It is wise to design to wire insulation to withstand the expected working voltage and a substantial safety margin.
See On use of enameled wire in transmitting baluns for more discussion.
I would avoid tinned copper wire (or any structure with lower conductivity coating), wires with low conductivity cores (eg CCS), highly stranded wires, unsolderable wires.
Dielectric loss for most good insulators at HF is approximately proportional to frequency, the length of line and applied voltage.
PVC insulated wires of used by lots of authors, but PVC dielectric loss tends to become significant towards 30MHz. In my experience, its LossFactor ranges widely and would seem to depend on pigments, plasticisers and fillers used in manufacture. It is not a good choice, especially at the high end of HF for 5 stacked cores (ie long line section) at 1500W on a high Z / unkown load.
Conductor loss is fairly straight forward, it includes skin effect and in the case where two conductors are very close together, proximity effect. Broadly, doubling wire diameter halves effective RF resistance, but this is modified by proximity effect.
Lets model a scenario:
Firstly let’s model it with a 50+j0Ω load, this is what you might expect if you measured the line section between Port 1 and Port 2 of a VNA.
1.6mm (#14) PVC twin 5 x FT240 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.001600 m 
Spacing  0.002500 m 
Velocity factor  0.800 
Loss tangent  5.000e3 
Frequency  30.000 MHz 
Twist rate  0 t/m 
Length  3.100 m 
Zload  50.00+j0.00 Ω 
Yload  0.020000+j0.000000 S 
Results  
Zo  97.93j0.22 Ω 
Velocity Factor  0.8000 
Twist factor  1.0000 
Rel permittivity  1.562 
R, L, G, C  7.362168e1, 4.098371e7, 4.027604e5, 4.273420e11 
Length  139.597 °, 0.387768 λ, 3.100000 m, 1.293e+4 ps 
Line Loss (matched)  0.154 dB 
Line Loss  0.182 dB 
Efficiency  95.89 % 
Zin  7.286e+1j4.992e+1 Ω 
Yin  0.00934021+j0.00639912 S 
VSWR(50)in, RL(50)in, MML(50)in  2.41, 7.660 dB 0.817 dB 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (source end)  5.648e2j3.076e1, 0.313∠100.4°, 10.097 dB, 1.91, 0.447 dB 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (load end)  3.240e1+j1.014e3, 0.324∠179.8°, 9.789 dB, 1.96, 0.482 dB 
V2/V1  3.543e1j5.677e1, 6.692e1∠122.0° 
I2/I1  1.083e+0j4.735e1, 1.182e+0∠156.4° 
I2/V1  7.087e3j1.135e2, 1.338e2∠122.0° 
V2/I1  5.416e+1j2.367e+1, 5.910e+1∠156.4° 
S11, S21 (50)  3.014e1j2.838e1, 6.223e1j6.382e1 
Let’s get s21(50) into polar form.
Because Zo is not 50+j0Ω, there is impedance transformation and Zin is 73j50Ω, s11=7.7dB s21=0.999dB. There are two contributions to s21:
Now let’s model it with a 2000Ω load to evaluate high voltage conditions.
1.6mm (#14) PVC twin 5 x FT240 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.001600 m 
Spacing  0.002500 m 
Velocity factor  0.800 
Loss tangent  5.000e3 
Frequency  30.000 MHz 
Twist rate  0 t/m 
Length  3.100 m 
Zload  2.000e+3+j0.000e+0 Ω 
Yload  5.000e4+j0.000e+0 S 
Results  
Zo  97.93j0.22 Ω 
Velocity Factor  0.8000 
Twist factor  1.0000 
Rel permittivity  1.562 
R, L, G, C  7.362168e1, 4.098371e7, 4.027604e5, 4.273420e11 
Length  139.597 °, 0.387768 λ, 3.100000 m, 1.293e+4 ps 
Line Loss (matched)  0.154 dB 
Line Loss  1.421 dB 
Efficiency  72.09 % 
Zin  1.604e+1+j1.159e+2 Ω 
Yin  0.00117212j0.00846881 S 
VSWR(50)in, RL(50)in, MML(50)in  20.13, 0.864 dB 7.439 dB 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (source end)  1.552e1+j8.611e1, 0.875∠79.8°, 1.160 dB, 15.00, 6.301 dB 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (load end)  9.066e1+j2.016e4, 0.907∠0.0°, 0.851 dB, 20.42, 7.496 dB 
V2/V1  1.298e+0j7.233e2, 1.300e+0∠176.8° 
I2/I1  6.217e3j7.577e2, 7.603e2∠94.7° 
I2/V1  6.490e4j3.617e5, 6.500e4∠176.8° 
V2/I1  1.243e+1j1.515e+2, 1.521e+2∠94.7° 
S11, S21 (50)  3.014e1j2.838e1, 6.223e1j6.382e1 
(transmission) Loss is now considerably higher than the base model at 1.4dB, 28% of input power is converted to heat due to conductor resistance and dielectric loss.
Now let’s model it with a 5Ω load to evaluate high current conditions.
1.6mm (#14) PVC twin 5 x FT240 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.001600 m 
Spacing  0.002500 m 
Velocity factor  0.800 
Loss tangent  5.000e3 
Frequency  30.000 MHz 
Twist rate  0 t/m 
Length  3.100 m 
Zload  5.00+j0.00 Ω 
Yload  0.200000+j0.000000 S 
Results  
Zo  97.93j0.22 Ω 
Velocity Factor  0.8000 
Twist factor  1.0000 
Rel permittivity  1.562 
R, L, G, C  7.362168e1, 4.098371e7, 4.027604e5, 4.273420e11 
Length  139.597 °, 0.387768 λ, 3.100000 m, 1.293e+4 ps 
Line Loss (matched)  0.154 dB 
Line Loss  1.231 dB 
Efficiency  75.31 % 
Zin  1.121e+1j8.120e+1 Ω 
Yin  0.00166894+j0.01208536 S 
VSWR(50)in, RL(50)in, MML(50)in  16.38, 1.062 dB 6.638 dB 
(transmission) Loss is now considerably higher than the base model at 1.2dB, 25% of input power is converted to heat due to conductor resistance and dielectric loss.
In summary, the high current and high voltage application both have much higher loss than the base 50 configuration that might be measured with a VNA (though it is not simply s21).
This design above is notable for having a very large number of turns on a 5xFT240 core stack. It is an unusual design and the long embedded transmission line with low grade PVC dielectric contributes to poor performance.
By contrast, a single FT240 with 11t of 1.6mm (#14) PTFE insulated wire and 2000Ω load would have transmission line loss closer to 0.1dB or 2% of input power converted to heat in the transmission line component.
The base design might be thought of as massive and capable of very high power, but whilst piling turns on might tend to reduce core losses, transmission line conductor and dielectric losses become more significant. Losing more than 25% of input power to transmission line loss with extreme loads on a 5x stack does not make for a kilowatt balun!
]]>In each case, a single wire is tested, one electrode to the wire and another being an alligator clip clipped onto the wire about 30mm from the end. This approximates a knife edge test which subjects the insulation to the highest electric field strength.
At the time of the test, temperature was 21° and relative humidity 65%. Whilst not extreme humidity, it is sufficient to degrade breakdown often giving rise of an arc over the surface of the wire to the cut end. For that reason, about 30mm of insulation is left clear at each end.
The source is 50Hz AC, so peak voltage is 1.414 times the RMS value shown on the instrument.
Above, the orange PTFE insulated wire on essentially a knife edge test at 6.2kV AC, 8.7kVpk. The conductor is stranded and 1.0mm diameter (#18), diameter over the insulation is 1.5mm, so about 0.25mm thickness PTFE.
Above, the red PTFE insulated wire on essentially a knife edge test at 6.2kV AC, 8.7kVpk. The conductor is stranded and 0.9mm diameter (#19), diameter over the insulation is 1.3mm, so about 0.2mm thickness PTFE.
I have used both these wires to wind ‘tuner baluns’ where the wire is subjected to quite high voltages. Whilst on the basis of these tests we might expect that the wire to wire withstand is more than 16kV, breakdown of other components like a balun input coax connector will set a practical limit of something less than 5kV. Breakdown is degraded by high humidity, and accumulated dirt on wires, connectors and terminals, so again tests of clean wire in low humidity need to be interpreted sensibly.
These wires can be obtained on eBay or Aliexpress for modest cost.
]]>The designer did report measurement at the choke looking into the feed line giving Z=493j740Ω @ 3.8MHz. There are questions about the validity / uncertainty of the measurement, but let’s take is as correct for the purpose of this discussion.
We can calculate the expected differential peak voltage at a given power level at the point where Z=493j740Ω.
Let’s use Python as a complex number calculator to calculate based on Z=493j740Ω.
>>> import math >>> import cmath >>> p=1500 >>> z=493740j >>> g=(1/z).real >>> '%0.2e'%g '6.24e04' >>> vpk=(2*p/g)**0.5 >>> '%0.2e'%vpk '2.19e+03'
Above, the solution of \(V_{pk}=\sqrt{\frac{2 p}{real(\frac1{z})}}\) gives the peak voltage as 2190V.
If your analyser / VNA gives the impedance as an equivalent parallel Rp  Xp, then the following might be more convenient using Rp.
>>> import math >>> import cmath >>> p=1500 >>> rp=1603.75 >>> vpk=(2*p*rp)**0.5 >>> '%0.2e'%vpk '2.19e+03'
Above, the solution of \(V_{pk}=\sqrt{2 p r_p}\) gives the peak voltage as 2190V.
Another simpler alternative is Calculate Vmax, Vmin, Imax, Imin for lossless line from Z (or Y or VSWR) and Zo.
Again, the answer is 2190Vpk.
Which ever way you get it, is 2190Vpk ok?
Given that the prototype balun was wound with enameled copper wire, you might think it should easily withstand 510kV wire to wire… but it would appear to have failed in this case at less than a third of that. The voltage withstand of the enamel insulation depends on many things, and can be degraded by sharp bends (eg around sharp core edges if present). In this case the core material was #31 which is not an extremely good insulator, so that might degrade things.
My experience having tested the withstand of some samples of enameled copper wire is that it not a good choice for over perhaps 1000Vpk for several reasons set out at On use of enameled wire in transmitting baluns.
It is very easy to damage enamel insulation with a single flashover, and thereafter the carbon track remaining in the enamel results in greatly reduced voltage withstand. The the arc might exist for milliseconds, that is sufficient to permanently damage the insulation… and therefore the whole choke.
The calculations given here are for differential voltage at the point where Zin is known. The differential voltage at other parts of the choke may be higher or lower… higher would be a bigger problem but it is not higher in this scenario.
]]>Sometimes the same dimensioned cores are available in both categories with different part numbers and possibly different prices, implying some real difference in behavior, eg 5943003801 and 2643803802 are both FT24043 sized cores.
Material datasheets often contain a note like this from the #43 datasheet:
Characteristic curves are measured on standard Toroids (18/10/6 mm) at 25°C and 10 kHz unless otherwise indicated. Impedance characteristics are measured on standard shield beads (3.5/1.3/6.0 mm) unless otherwise indicated.
I sought to clarify my interpretation of this clause by asking Fairrite …whether the published material permeability curves / tables apply to suppression product. Can I use the published permeability curves / tables to predict inductor impedance reliably for suppression products?
Fairrite’s Michael Arasim advised…
Yes the published permeability curve can be used to predict impedance. There will be some variance in the shape of the curve due to individual part size as well as process and material variation. The level of this variance will change depending on the individual material but, the curves themselves are all produced using the same sized toroidal core for each material. One thing of note; Inductive rated parts in theory will be controlled more tightly to adhere to the complex permeability curve since their acceptance criteria is generally going to be inductance and loss factor based. Impedance rated parts are accepted based on a minimum impedance at select test frequencies. Since the impedance is a complex value influenced by both the inductive and resistive components of the complex permeability; In theory you could see more variance in each component and still hit the impedance rating. Prior to production, all lots of our materials are screened to ensure that they will adhere (within a tolerance) to the published material data.
So, the following notes and tools are applicable to Fairrite inductive and suppression products, but one might expect more variation in the real and imaginary components of impedance with suppression products, and loss in cores used for transformers.
My own experience is that the difference is not huge, but it explains the somewhat wider variation observed with suppression products and the need to verify designs by measurement.
]]>The calculator has been revised to include 45° chamfers of a specified length on all four corners. If the chamfer angle differs, the error is very small in the range 3060°. If the corners are radiused, use the radius as the chamfer length, the error is very small.
We do not need to obsess over these errors as they will usually be dwarfed by manufacturing tolerances.
The calculation of ΣA/l for the sharp corner model is fairly simple.
\(\int _{ir}^{or}\frac{w}{2 \pi r}dr\)
\(=\frac{w}{2 \pi }\left(\ln \left(or\right)\ln \left(ir\right)\right)\)
To implement the chamfer adjustment, the ΣA/l component of the missing material is calculated.
Firstly the inner chamfers (which are simpler).
\(\int _{ir}^{ir+cr}\:\frac{rir}{2 \pi r}dr\)
\(=\frac{1}{2\pi}\left(crir\left(\ln \left(ir+cr\right)\ln \left(ir\right)\right)\right)\)
Then the outer chamfers are calculated.
\(\int _{orcr}^{or} \frac{\left(r\left(orcr\right)\right)}{2 \pi r}dr\)
\(=\frac{1}{2\pi }\left(cr\left(\ln \left(or\right)\ln \left(cr+or\right)\right)or\left(\ln \left(or\right)\ln \left(cr+or\right)\right)+cr\right)\)
The final ΣA/l is the first quantity less the two missing components.
Implemented in javascript in the calculator…
aol=width/2/pi*Math.log(od/id)*1e3; aol=aol(crir*(Math.log(ir+cr)Math.log(ir)))/(2*pi)*1e3*2; aol=aol(cr*(Math.log(or)Math.log(cr+or))or*(Math.log(or)Math.log(cr+or))+cr)/(2*pi)*1e3*2;
Let’s look at some examples.
Above is a calculation for the popular FT24043. Without the chamfer adjustment, ΣA/l would be 0.001091 (which is the value given by Fairrite in the datasheet), adjusting for chamfer the reduction is 2.5%.
Above is an example calculation of ΣA/l and Al. the calculated ΣA/l is less than 1% less than if the chamfer were ignored. The difference may be greater on some cores, especially very small cores.
Above is an example calculation for a very small core with radiused corners. The chamfer approximation reduces ΣA/l and Al by about 2%… again the manufacturing tolerances dwarf the adjustment.
Sizing the adjustment is an accurate way to determine if it is significant or not, and 2% accuracy does not have a lot of application to ferrites with 20% tolerance… but it does become more important when trying to characterise the ferrite material based on core dimensions.
]]>Above, the core is 35x21x13mm, a mid sized core, two used in my redesign of a commercial balun and implemented by VK4MQ . The mid size limits dissipation, but compactness can be an advantage. The cores sell for less than $4.00 per core and are readily available in Australia.
The core is almost certainly made in China, and Jaycar does not publish complex permeability curves for the material, but above are my measured characteristics over HF. The Chinese factor does raise questions about continued supply of consistent quality product.
Above is a plot of Fairrite’s complex permeability for #31 ferrite material.
If you compare the two, they are not identical, but are very similar and you could conclude that applications where #31 is a good material selection would be well served by L15. Notably #31 is a MnZn ferrite and the L15 appears to be a NiZn ferrite based on its very high resistivity.
It is interesting to observe the fashions in online discussions of the best balun material
that the current fashion amongst online experts is #31. #31 is certainly a good candidate for applications with emphasis on lower HF, but its suitability for a specific applications needs also to consider other factors like its loss.
For the same reason, Jaycar’s LO1238 using L15 material may be quite suitable to those type of applications.
]]>In respect of the first part, inductance \(L=\frac{\phi(i)}{i}\) so if the windings are equal, half the total current flows in each winding and each contributes flux due to i/2, total current is i, total flux is twice that due to i/2, so the inductance of the parallel equal windings is the same as if i flowed in a single winding, ie L of the combination is the same as the inductance of each of the equal windings alone.
In the second case, if there is zero flux leakage, it is true that the inductance of the combination of series opposed equal windings is zero. In the case of the cores used for common mode chokes flux leakage varies from less than 1% to over 50%… so the general statement is a bit naive in this application.
Both of these are very easily demonstrated by simple experiment. In fact a measurement of L with series aiding and series opposing connection is a classic way to find the flux coupling factor.
A few years after the ‘information’ was posted, it has not been questioned much less called out as wrong. So given that the understanding of inductance seems lacking… what credibility does the article have when it includes this?
]]>