Inductors in Series: Understanding Equivalent Inductance and Mutual Inductance

When two inductors, say \(L_1\) and \(L_2\), are connected in series, their combined or equivalent inductance (\(L_{eq}\)) depends on whether their magnetic fields aid or oppose each other. This interaction is described by a property called mutual inductance (\(M\)).

Inductance Formulas for Series Connection

For two inductors connected in series, the equivalent inductance is calculated using one of two formulas, depending on the coupling:

• Series Aiding Connection: In this configuration, the magnetic fields produced by the two inductors add up, increasing the total inductance. The formula is: \[L_{eq} = L_1 + L_2 + 2M\] Here, \(L_1\) and \(L_2\) are the self-inductances of the individual coils, and \(M\) is the mutual inductance between them.
• Series Opposing Connection: In this configuration, the magnetic fields produced by the two inductors partially cancel each other out, decreasing the total inductance. The formula is: \[L_{eq} = L_1 + L_2 - 2M\] Again, \(L_1\) and \(L_2\) are self-inductances, and \(M\) is the mutual inductance.

Calculating Mutual Inductance Step-by-Step

Let's apply these concepts to the given problem. We are provided with the following information:

• Self-inductance of the first inductor, \(L_1 = 20 \text{ mH}\)
• Self-inductance of the second inductor, \(L_2 = 40 \text{ mH}\)
• Equivalent inductance of the series connection, \(L_{eq} = 50 \text{ mH}\)

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

First, let's consider the scenario where the inductors are connected in series aiding. If they were series aiding, the equivalent inductance would be:

\[L_{eq} = L_1 + L_2 + 2M\] \[50 \text{ mH} = 20 \text{ mH} + 40 \text{ mH} + 2M\] \[50 = 60 + 2M\] \[2M = 50 - 60\] \[2M = -10 \text{ mH}\] Since mutual inductance (\(M\)) cannot be negative (as it represents a physical interaction that transfers energy), this indicates that the inductors are not connected in series aiding.

Therefore, the inductors must be connected in series opposing. Using the formula for series opposing connection:

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

Now, substitute the given values into this equation:

\[50 \text{ mH} = 20 \text{ mH} + 40 \text{ mH} - 2M\]

Combine the self-inductances:

\[50 = 60 - 2M\]

To find \(2M\), rearrange the equation:

\[2M = 60 - 50\] \[2M = 10 \text{ mH}\]

Finally, calculate \(M\) by dividing by 2:

\[M = \frac{10 \text{ mH}}{2}\] \[M = 5 \text{ mH}\]

Summary of Inductor Calculation

The mutual inductance between the two coils is \(5 \text{ mH}\). This calculation confirms that for the equivalent inductance to be less than the sum of individual inductances (\(20 \text{ mH} + 40 \text{ mH} = 60 \text{ mH}\)), the inductors must be connected in a series opposing manner.