If I have a set of coils with 39,5mh each and when I connect them in series their inductance is 148,6mh what is the mutual inductance and what would be the inductance when I connect them in series with fields opposing ? Their resistance is 18 ohms each and it’s air core

To find the mutual inductance (\( M \)) of the coils and the inductance when connected in series with fields opposing, we can use the following equations.

### Inductance with Fields in Series (Aiding):

When two inductors are connected in series with their fields aiding, the total inductance (\( L_{aiding} \)) is given by:

\[ L_{aiding} = L_1 + L_2 + 2M \]

where:

- \( L_1 \) and \( L_2 \) are the inductances of the individual coils.

- \( M \) is the mutual inductance.

Given:

\[ L_{aiding} = 148.6 \text{ mH} \]

\[ L_1 = L_2 = 39.5 \text{ mH} \]

Plugging these into the formula, we get:

\[ 148.6 \text{ mH} = 39.5 \text{ mH} + 39.5 \text{ mH} + 2M \]

\[ 148.6 \text{ mH} = 79 \text{ mH} + 2M \]

\[ 2M = 148.6 \text{ mH} - 79 \text{ mH} \]

\[ 2M = 69.6 \text{ mH} \]

\[ M = \frac{69.6 \text{ mH}}{2} \]

\[ M = 34.8 \text{ mH} \]

So, the mutual inductance (\( M \)) is 34.8 mH.

### Inductance with Fields Opposing:

When the fields oppose, the total inductance (\( L_{opposing} \)) is given by:

\[ L_{opposing} = L_1 + L_2 - 2M \]

Using the values we have:

\[ L_{opposing} = 39.5 \text{ mH} + 39.5 \text{ mH} - 2 \times 34.8 \text{ mH} \]

\[ L_{opposing} = 79 \text{ mH} - 69.6 \text{ mH} \]

\[ L_{opposing} = 9.4 \text{ mH} \]

Therefore, the inductance when the coils are connected in series with their fields opposing is 9.4 mH.

Each inductance has a value of 86mh and in parallel the 3 have 39,5 what is the mutual inductance please?

Given the following data:

1. Each coil has an inductance \( L = 86 \text{ mH} \).

2. When three identical coils are connected in parallel, the total inductance is \( L_{parallel3} = 39.5 \text{ mH} \).

We need to determine the mutual inductance \( M \) between the coils.

### Inductors in Parallel:

For three identical inductors in parallel, considering mutual inductance \( M \), the formula for the equivalent inductance \( L_{parallel3} \) is:

\[ \frac{1}{L_{parallel3}} = \frac{1}{L + M} + \frac{1}{L + M} + \frac{1}{L + M} \]

Given:

\[ L = 86 \text{ mH} \]

\[ L_{parallel3} = 39.5 \text{ mH} \]

### Mutual Inductance with Parallel Configuration:

\[ \frac{1}{L_{parallel3}} = \frac{3}{L + M} \]

So:

\[ L_{parallel3} = \frac{L + M}{3} \]

Given \( L_{parallel3} = 39.5 \text{ mH} \):

\[ 39.5 = \frac{86 + 2M}{3} \]

Solving for \( M \):

\[ 3 \times 39.5 = 86 + 2M \]

\[ 118.5 = 86 + 2M \]

\[ 118.5 - 86 = 2M \]

\[ 32.5 = 2M \]

\[ M = \frac{32.5}{2} \]

\[ M = 16.25 \text{ mH} \]

Therefore, the mutual inductance \( M \) is 16.25 mH.

The coupling coefficient (\( k \)) is a measure of how effectively two inductors are magnetically coupled. It is defined as:

\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Given that each coil has an inductance \( L_1 = L_2 = 86 \text{ mH} \), and the mutual inductance \( M \) has been calculated as \( 16.25 \text{ mH} \):

\[ k = \frac{16.25}{\sqrt{86 \cdot 86}} \]

First, calculate the product of \( L_1 \) and \( L_2 \):

\[ L_1 \cdot L_2 = 86 \cdot 86 = 7396 \text{ mH}^2 \]

Next, take the square root of the product:

\[ \sqrt{7396} = 86 \text{ mH} \]

Now, calculate the coupling coefficient \( k \):

\[ k = \frac{16.25}{86} \]

\[ k \approx 0.189 \]

Therefore, the coupling coefficient \( k \) in this case is approximately 0.189.

And in the case of having them in series 148mh and in parallel 9,4mh what is the coupling being the mutual inductance 34,8

To find the coupling coefficient (\( k \)) given the series and parallel inductance values, we will use the provided mutual inductance (\( M \)).

### Given:

- Series inductance: \( L_{series} = 148 \text{ mH} \)

- Parallel inductance: \( L_{parallel} = 9.4 \text{ mH} \)

- Mutual inductance: \( M = 34.8 \text{ mH} \)

### Inductance of Each Coil:

From the series inductance formula, we know:

\[ L_{series} = L_1 + L_2 + 2M \]

Given \( L_1 = L_2 = L \) (assuming identical coils):

\[ 148 = L + L + 2 \times 34.8 \]

\[ 148 = 2L + 69.6 \]

\[ 2L = 148 - 69.6 \]

\[ 2L = 78.4 \]

\[ L = 39.2 \text{ mH} \]

So, the inductance of each coil is \( L = 39.2 \text{ mH} \).

### Coupling Coefficient:

The coupling coefficient (\( k \)) is given by:

\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Since \( L_1 = L_2 = L \):

\[ k = \frac{34.8}{\sqrt{39.2 \cdot 39.2}} \]

\[ k = \frac{34.8}{39.2} \]

\[ k = 0.887 \]

Therefore, the coupling coefficient (\( k \)) is 0.887.

So the result was exact as I designed to be and expected

The coils have a low coupling between them allowing to lower the inductance when connect them in parallel

The coupling between them side by side is high as I expected 88% this couples the resonance together

Is possible that the screw inside holding it may affect some the readings but is just to get the idea..