Inductance of a coil is calculated according to the attached formula (1), where Mij is the mutual inductance between loops. (Mutual inductance can be calculated using different methods, as example using elliptic integrals of first and second order.)
My question is how to calculate the inductance if the loops are connected in parallel instead of in series.
Could I use the second formula?
I obtained it using formulas written in the attached document.
Data Preliminary INDUCTANCE CALCULATION OF PARALLEL CONNECTED COI...
Visit http://www.electronics-tutorials.ws/inductor/parallel-inductors.html and see the section entitled "Mutually Coupled Inductors in Parallel"
My inductors are stacked end to end, and turn on separately.
Thank you!, your equation and mine results "almost" the same for 2 inductors! :
Matrix=
|L1 L12|
|L12 L2|
inverse= 1/det x
|L2 -L12|
|-L12 L1|
so 1/sum(inverse members) is:
L=(L1L2-L12^2)/(L1+L2-2L12)
It would be a good idea that somebody make a test to add him in a paper
Here is the full detail of inductors in parallel coil .
Inductors are said to be connected together in “Parallel” when both of their terminals are respectively connected to each terminal of the other inductor or inductors. The voltage drop across all of the inductors in parallel will be the same. Then, Inductors in Parallel have a Common Voltage across them and in our example below the voltage across the inductors is given as:
VL1 = VL2 = VL3 = VAB …etc
In the following circuit the inductors L1, L2 and L3 are all connected together in parallel between the two points A and B.
Inductors in Parallel Circuit
inductors in parallel
In the previous series inductors tutorial, we saw that the total inductance, LT of the circuit was equal to the sum of all the individual inductors added together. For inductors in parallel the equivalent circuit inductance LT is calculated differently.
The sum of the individual currents flowing through each inductor can be found using Kirchoff’s Current Law (KCL) where, IT = I1 + I2 + I3 and we know from the previous tutorials on inductance that the self-induced emf across an inductor is given as: V = L di/dt
Then by taking the values of the individual currents flowing through each inductor in our circuit above, and substituting the current i for i1 + i2 + i3 the voltage across the parallel combination is given as:
inductor currents
By substituting di/dt in the above equation with v/L gives:
derative reduction
We can reduce it to give a final expression for calculating the total inductance of a circuit when connecting inductors in parallel and this is given as:
Parallel Inductor Equation
parallel inductance
Here, like the calculations for parallel resistors, the reciprocal ( 1/Ln ) value of the individual inductances are all added together instead of the inductances themselves. But again as with series connected inductances, the above equation only holds true when there is “NO” mutual inductance or magnetic coupling between two or more of the inductors, (they are magnetically isolated from each other). Where there is coupling between coils, the total inductance is also affected by the amount of coupling.
This method of calculation can be used for calculating any number of individual inductances connected together within a single parallel network. If however, there are only two individual inductors in parallel then a much simpler and quicker formula can be used to find the total inductance value, and this is:
parallel inductors
One important point to remember about inductors in parallel circuits, the total inductance ( LT ) of any two or more inductors connected together in parallel will always be LESS than the value of the smallest inductance in the parallel chain.
Dear prof. Bhupendra, I should forget Kirchhoff with coupled inductances, it can be seen more easily using matrices.
V0=M00*dI0 + M01*dI1 + M02*dI2
V1=M10*dI0 + M11*dI1 + M12*dI2
V2=M20*dI0 + M21*dI1 + M22*dI2
Where:
Mij=Mji
V0=voltage between inductor-0 terminals
dI0= -derivative of current through inductor-0 with respect time
When inductors are in parallel you have:
V0=V1=V2, and dIt=dI0 + dI1 + dI2
When inductors are connected serially then:
Vt=V0+V1+V2
dIt = dI0 = dI1 = dI2
Note: I recommend to upload drawings as files or upload all together as a file
- It would be a good thing to have a equation editor in the forum -
Then to obtain individual dIx/dt it can be done by using matrices:
[Vi]=[Mij]*[dIj/dt]
[dIj/dt]=[Mij]´ * [Vi]
Note: [Mij]´ is the inverse of matrix [Mij]
No, I haven't, I program usually using C++ because it is much faster to make simulations, but I am not very happy with results, as example perpendicular coils give 0 mutual inductance that is totally wrong accordingly tests if the coils are not in the same axis
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