**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:

V_{L1} = V_{L2} = V_{L3} = V_{AB} …etc

In the following circuit the inductors L_{1}, L_{2} and L_{3} are all connected together in parallel between the two points A and B.

In the previous series inductors tutorial, we saw that the total inductance, L_{T} of the circuit was equal to the sum of all the individual inductors added together. For inductors in parallel the equivalent circuit inductance L_{T} is calculated differently.

The sum of the individual currents flowing through each inductor can be found using Kirchoff’s Current Law (KCL) where, I_{T} = I_{1} + I_{2} + I_{3} 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 i_{1} + i_{2} + i_{3} the voltage across the parallel combination is given as:

By substituting di/dt in the above equation with v/L gives:

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:

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:

One important point to remember about inductors in parallel circuits, the total inductance ( L_{T} ) 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.

Three inductors of 60mH, 120mH and 75mH respectively, are connected together in a parallel combination with no mutual inductance between them. Calculate the total inductance of the parallel combination in millihenries.

When inductors are connected together in parallel so that the magnetic field of one links with the other, the effect of mutual inductance either increases or decreases the total inductance depending upon the amount of magnetic coupling that exists between the coils. The effect of this mutual inductance depends upon the distance apart of the coils and their orientation to each other.

Mutually connected inductors in parallel can be classed as either “aiding” or “opposing” the total inductance with parallel aiding connected coils increasing the total equivalent inductance and parallel opposing coils decreasing the total equivalent inductance compared to coils that have zero mutual inductance.

Mutual coupled parallel coils can be shown as either connected in an aiding or opposing configuration by the use of polarity dots or polarity markers as shown below.

The voltage across the two parallel aiding inductors above must be equal since they are in parallel so the two currents, i_{1} and i_{2} must vary so that the voltage across them stays the same. Then the total inductance, L_{T} for two parallel aiding inductors is given as:

Where: 2M represents the influence of coil L _{1} on L _{2} and likewise coil L _{2} on L _{1}.

If the two inductances are equal and the magnetic coupling is perfect such as in a toroidal circuit, then the equivalent inductance of the two inductors in parallel is L as L_{T} = L_{1} = L_{2} = M. However, if the mutual inductance between them is zero, the equivalent inductance would be L ÷ 2 the same as for two self-induced inductors in parallel.

If one of the two coils was reversed with respect to the other, we would then have two parallel opposing inductors and the mutual inductance, M that exists between the two coils will have a cancelling effect on each coil instead of an aiding effect as shown below.

Then the total inductance, L_{T} for two parallel opposing inductors is given as:

This time, if the two inductances are equal in value and the magnetic coupling is perfect between them, the equivalent inductance and also the self-induced emf across the inductors will be zero as the two inductors cancel each other out.

This is because as the two currents, i_{1} and i_{2} flow through each inductor in turn the total mutual flux generated between them is zero because the two flux’s produced by each inductor are both equal in magnitude but in opposite directions.

Then the two coils effectively become a short circuit to the flow of current in the circuit so the equivalent inductance, L_{T} becomes equal to ( L ± M ) ÷ 2.

Two inductors whose self-inductances are of 75mH and 55mH respectively are connected together in parallel aiding. Their mutual inductance is given as 22.5mH. Calculate the total inductance of the parallel combination.

Calculate the equivalent inductance of the following inductive circuit.

Calculate the first inductor branch L_{A}, (Inductor L_{5} in parallel with inductors L_{6} and L_{7})

Calculate the second inductor branch L_{B}, (Inductor L_{3} in parallel with inductors L_{4} and L_{A})

Calculate the equivalent circuit inductance L_{EQ}, (Inductor L_{1} in parallel with inductors L_{2} and L_{B})

Then the equivalent inductance for the above circuit was found to be: 15mH.

As with the resistor, inductors connected together in parallel have the same voltage, V across them. Also connecting together inductors in parallel decreases the effective inductance of the circuit with the equivalent inductance of “N” inductors connected in parallel being the reciprocal of the sum of the reciprocals of the individual inductances.

As with series connected inductors, mutually connected inductors in parallel are classed as either “aiding” or “opposing” this total inductance depending whether the coils are cumulatively coupled (in the same direction) or differentially coupled (in opposite direction).

Thus far we have examined the inductor as a pure or ideal passive component. In the next tutorial about Inductors, we will look at non-ideal inductors that have real world resistive coils producing the equivalent circuit of an inductor in series with a resistance and examine the time constant of such a circuit.

Error! Please fill all fields.

Thanks–very illuminating. I have a fiberglass box containing 3 pair of inductors located at the feedpoint of an 80m dipole to gain coverage of the entire band in 8 segments. One pair’s inductance is about 0.8 uH for each coil. a second pair’s is about 1.6 uH and a third pair’s is about 3.2 uH. 3 DPST relays were used short out or activate each pair. If 3 relays are open, 3 coils on each side of the relays are connected in series. The coils are set end to end with end to end spacing of about a coil diameter +/- and wound the same direction to aid each other if there is mutual coupling. My concern is: When an adjacent coil is shorted out, is an active coil next to it adversely affected due to mutual coupling such as a leaky transformer with the secondary shorted out?

If two or more coils are mutually coupled and one becomes faulty, then yes it will affect the other coils as the linking flux is different.

Series Inductors

Tell about series adding inductor n oppsing wt formula plz

Proof:Mutual aided coupling

I=i1+i2 parallel=>voltage equal v=v

L1 (di1)/dt + M (di2)/dt = L2 (di2)/dt + M (di1)/dt

=> i2 = i1*(L1– M)/ (L2– M) put this in

I= i1+ i2 = i1*(1+(L1– M)/ (L2– M))= i1 * (L1+L2-2M)/L2-M)

To find i1

v = L1 (di1)/dt) + M (di2)/dt

v = L2 (di2)/dt + M (di1)/dt

eliminate i2 by Gauss elimination method.

di1/dt = (L2– M)/(L1L2-M^2) * V

i1 = (L2– M)/(L1L2-M^2) * ∫ V dt

Put this i1 in eqation of I total obtained above.We get

I= (L1+L2-2M)/(L1L2-M^2) * ∫ V dt

1/L eq * ∫ V dt (which is I total) = 1/((L1L2-M^2) /(L1+L2-2M))* ∫ V dt

observe L eq = (L1L2-M^2) /(L1+L2-2M)

pls explain how the mutual inductance value becomes 22.5?

Please read the example description, the answer is there.

It is good

how we can proof equivalent inductance in case of two parallel inductance affected from mutual inductance : Leq = L1 . L 2 – m2 / L 1 + L 2 – m

Is wheat stone bridge concept applicable in case of inductors? pls give me the answer in the provided email addredd ….thnx BTW it is dwaipayanroy0088@gmail.com…..

Its a bit pointless, but yes it would be possible to use inductors with an AC supply instead of resistors. The Inductive Reactance, XL of the inductors would give the output AC signal voltage.

why my comment not published? moderated?

Due to the high levels of spam, links and general rubbish that people want to post, ALL comments are moderated for approval first. You have to wait.

hmm… I have seen the formula to calculate parallel inductors with mutual inductance involved, you have given in many sites. I doubt this formula is correct. Of course Banter’s question is valid, but nobody could answer it. That’s because they use a wrong formula. Based on my physics and general knowledge, the correct formula is

1/L = 1/(L1 + M) + 1/(L2 +M) – aiding

1/L = 1/(L1 – M) + 1/(L2 – M) – opposing

If you were to work this out algebraically, you may get

L = (L1L2 + L1M + L2M + M^2) / (L1 + L2 + 2M) – aiding

But the given formula is something like

L = (L1L2 – M^2) / (L1 + L2 – 2M)

I wonder how this formula is derived.

I think there is a slighly better way to present the formula for inductors in parallel w/no mutual inductance for Example 1. It is really a matter of being extra clear for noobs like me!

To avoid confusion, may I suggest instead of this:

Ltotal = 1/L1 + 1/L2 + 1/L3

use this instead:

` 1

Ltotal = ——————————

` (1/L1 + 1/L2 + 1/L3)

` 1

` Ltotal = ———————————————– = 26mH

` (1/60mH + 1/120mH + 1/75mH)

Sorry for the tick marks, but I wanted everything to line up correctly. I am not sure that it will…

I hope this is useful.

Hello Robert, Yes you are right, it could be better explained and I have amended the example. Thanks. 🙂