Mutual Inductance
Introduction: This article focuses on the situation where two circuit elements are linked by a magnetic field.
The Essentials
Mutual inductance in a circuit involves 2 inductors and the magnetic field. We have gone over the self-inductance of an inductor, which is just the amount of henrys in an inductor. Mutual inductance is shared between the two inductors and induces a current through one inductor based on the current through the other. Figure 1. shows a schematic of all necessary details for a mutually inductive circuit.
Figure 1: Mutual Inductance Explanation
There are a couple of details to mention in the schematic explanation. You will notice dots around the mutually inductive inductors. This dot convention describes the polarity of the voltage across both inductors. The windings of inductors is crucial to defining the induced current through one another.
A general rule of thumb is this: when the current through the left inductor "enters" its dot, a positive voltage is induced at the dot of the other inductor, and when the same current "leaves" its dot, a negative voltage is induced at the dot of the other inductor. The textbook contains in-depth analysis on the dot markings of mutually inductive inductors.
Knowing this now, the easiest way to analyze the circuit is through mesh analysis. If we're to include mutual inductance in this equation, we would get the following:
\[ -v_g + R_1i_1 + L_1\frac{di_1}{dt} - M\frac{di_2}{dt} = 0\]
\[ i_2R_2 + L_2\frac{di_2}{dt} - M\frac{di_1}{dt} = 0\]
At times, we may have a circuit described by a coupling coefficient, which is shown as \( k \). Equation 1 shows how to calculate the mutual inductance given two inductance values and a coupling coefficient.
When it comes to applying mutual inductance, one applicable device that uses this is transformers. Transformers convert power from one form to another with the help of coiled wires and magnetic fields. At the bottom of this article is a link explaining the details of transformers and using mutual inductance.
Example
Let's do a simple example to get a feel for mutual inductance. Figure 2. has an example of a circuit paired by mutual inductance.
Figure 2: Mutual Inductance Example Problem
For this, we'll practice just writing the mesh current equations. We recognize that \( i_s \) describes the current through the first mesh, and in many cases that will be given to us, so we are only concerned about the current through the second mesh. Since the current through mesh 1 is traveling clockwise, we assume the current in mesh 2 is traveling counter-clockwise. Looking at this circuit and applying the logic in the essentials section, we get the following:
\[ i_2R_L + L_2\frac{di_2}{dt}-M\frac{di_1}{dt} = 0\]
Practice
- Using Figure 3. and given that \( R_1 = 1.3 \, \text{k}\Omega \), \( R_2 = 4.5 \, \text{k}\Omega \), \( R_3 = 2.3 \, \text{k}\Omega \), \( L_1 = 45 \, \text{mH} \), \( L_2 = 60 \, \text{mH} \), and \( M = 25 \, \text{mH} \), come up with mesh current equations to solve for each mesh.
Figure 3: Problem 1
- Using Figure 4. and given that \( R_1 = 2.6 \, \text{k}\Omega \), \( R_2 = 8.9 \, \text{k}\Omega \), \( R_3 = 6.7 \, \text{k}\Omega \), \( L_1 = 450 \, \text{mH} \), \( L_2 = 760 \, \text{mH} \), and \( M = 850 \, \text{mH} \), come up with mesh current equations to solve for each mesh.
Figure 4: Problem 2
- Using Figure 5. and given that \( R_1 = 8.9 \, \text{k}\Omega \), \( R_2 = 9.7 \, \text{k}\Omega \), \( R_3 = 10.2 \, \text{k}\Omega \), \( R_4 = 5.6 \, \text{k}\Omega \) \( L_1 = 570 \, \text{mH} \), \( L_2 = 980 \, \text{mH} \), and \( M = 480 \, \text{mH} \), come up with mesh current equations to solve for each mesh.
Figure 5: Problem 3
Solutions:
Problem 1 Solution.) There are 4 mesh current equations:
\[i_1 = I_s\]
\[ 1.3 \cdot10^3(i_2-i_1)+45\cdot10^{-3}\frac{di_2}{dt} - 25 \cdot 10^{-3}\frac{d}{dt}(i_3-i_4) = 0 \]
\[ 45 \cdot 10^{-3}(i_3-i_2) + 25\cdot10^{-3}\frac{d}{dt}(i_3-i_4) + 4.5 \cdot 10^{-3}(i_2)+60 \cdot 10^{-3}\frac{d}{dt}(i_3-i_4) + 25 \cdot 10^{-3} \frac{d}{dt}(i_3-i_2) = 0 \]
\[ 60 \cdot 10^{-3}(i_4-i_3) - 25 \cdot 10^{-3}\frac{d}{dt}(i_2-i_3) + 2.3\cdot10^{-3}i_4 = 0 \]
Problem 2 Solution.) There are 3 mesh current equations:
\[ 8.9\cdot10^{-3}(i_1-i_2) + 450 \cdot 10^{-3}\frac{d}{dt}(i_1-i_3) + 850 \cdot 10^{-3}\frac{d}{dt}(i_2 - i_3) = 0\]
\[ 2.6 \cdot 10^3 (i_2) + 760 \cdot 10^{-3}\frac{d}{dt}(i_2-i_3) - 850 \cdot 10^{-3} \frac{d}{dt}(i_3-i_1) + 8.9 \cdot 10^3(i_2-i_1) = 0\]
\[ 450 \cdot 10^{-3}\frac{d}{dt}(i_3-i_1) - 850 \cdot 10^{-3}\frac{d}{dt}(i_3-i_2) + 760 \cdot 10^{-3}\frac{d}{dt}(i_3-i_2)+850 \cdot 10^{-3}\frac{d}{dt}(i_2-i_3) + 6.7 \cdot 10^{3}i_3 = 0\]
Problem 3 Solution.) There are 4 mesh current equations:
\[ V_s = 8.9\cdot10^3(i_1-i_2) \]
\[ 8.9\cdot10^3(i_2-i_1) + 570\cdot10^{-3}\frac{di_2}{dt} - 480 \cdot 10^{-3}\frac{d}{dt}(i_3-i_2) + 980 \cdot 10^{-3}\frac{d}{dt} (i_2-i_3) - 480 \cdot 10^{-3}\frac{di_2}{dt} = 0\]
\[ 980 \cdot 10^{-3}\frac{d}{dt}(i_3-i_2) - 480 \cdot 10^{-3}\frac{di_2}{dt} + 10.2 \cdot 10^3(i_3-i_4) + 5.6 \cdot 10^3(i_3) = 0 \]
\[ 9.7 \cdot 10^3(i_4) + 10.2 \cdot 10^3(i_4 - i_3) = 0 \]