# Networks with Magnetically Coupled Coils   Whatsapp  For multiloop networks with magnetically coupled coils, the mesh analysis approach is most frequently applied. A firm understanding of the dot convention discussed earlier should make the writing of the equations quite direct and free of errors. Before writing the equations for any particular loop, first determine whether the mutual term is positive or negative, keeping in mind that it will have the same sign as that for the other magnetically coupled coil. For the two-loop network of Fig. 1, for example, the mutual term has a positive sign since the current through each coil leaves the dot. For the primary loop, Fig. 1: Applying mesh analysis to magnetically coupled coils.
$$\mathbf{E}_{1}-\mathbf{I}_{1} \mathbf{Z}_{1}-\mathbf{I}_{1} \mathbf{Z}_{L_{1}}-\mathbf{I}_{2} \mathbf{Z}_{m}-\mathbf{Z}_{2}\left(\mathbf{I}_{1}-\mathbf{I}_{2}\right)=0$$
where $M$ of $\mathbf{Z}_{m}=\omega M \angle 90^{\circ}$ is positive, and
$$\mathbf{I}_{1}\left(\mathbf{Z}_{1}+\mathbf{Z}_{L_{1}}+\mathbf{Z}_{2}\right)-\mathbf{I}_{2}\left(\mathbf{Z}_{2}-\mathbf{Z}_{m}\right)=\mathbf{E}_{1}$$
Note in the above that the mutual impedance was treated as if it were an additional inductance in series with the inductance $L_{1}$ having a sign determined by the dot convention and the voltage across which is determined by the current in the magnetically coupled loop.
For the secondary loop,
$$\begin{array}{ll} & -\mathbf{Z}_{2}\left(\mathbf{I}_{2}-\mathbf{I}_{1}\right)-\mathbf{I}_{2} \mathbf{Z}_{L_{2}}-\mathbf{I}_{1} \mathbf{Z}_{m}-\mathbf{I}_{2} \mathbf{Z}_{3}=0 \\\text { or } & \mathbf{I}_{2}\left(\mathbf{Z}_{2}+\mathbf{Z}_{L_{2}}+\mathbf{Z}_{3}\right)-\mathbf{I}_{1}\left(\mathbf{Z}_{2}-\mathbf{Z}_{m}\right)=0\end{array}$$ Fig. 2: Applying mesh analysis to a network with two magnetically coupled coils.
For the network of Fig. 2, we find a mutual term between $L_{1}$ and $L_{2}$ and $L_{1}$ and $L_{3}$, labeled $M_{12}$ and $M_{13}$, respectively.
For the coils with the dots $\left(L_{1}\right.$ and $\left.L_{3}\right)$, since each current through the coils leaves the dot, $M_{13}$ is positive for the chosen direction of $I_{1}$ and $I_{3}$. However, since the current $I_{1}$ leaves the dot through $L_{1}$, and $I_{2}$ enters the dot through coil $L_{2}, M_{12}$ is negative. Consequently, for the input circuit,
$$\mathbf{E}_{1}-\mathbf{I}_{1} \mathbf{Z}_{1}-\mathbf{I}_{1} \mathbf{Z}_{L_{1}}-\mathbf{I}_{2}\left(-\mathbf{Z}_{m_{12}}\right)-\mathbf{I}_{3} \mathbf{Z}_{m_{13}}=0$$or$$\mathbf{E}_{1}-\mathbf{I}_{1}\left(\mathbf{Z}_{1}+\mathbf{Z}_{L_{1}}\right)+\mathbf{I}_{2} \mathbf{Z}_{m_{12}}-\mathbf{I}_{3} \mathbf{Z}_{m_{13}}=0$$
For loop 2 ,
$$\begin{array}{r}-\mathbf{I}_{2} \mathbf{Z}_{2}-\mathbf{I}_{2} \mathbf{Z}_{L_{2}}-\mathbf{I}_{1}\left(-\mathbf{Z}_{m_{12}}\right)=0 \\-\mathbf{I}_{1} \mathbf{Z}_{m_{12}}+\mathbf{I}_{2}\left(\mathbf{Z}_{2}+\mathbf{Z}_{L_{2}}\right)=0\end{array}$$
and for loop 3,
$$-\mathbf{I}_{3} \mathbf{Z}_{3}-\mathbf{I}_{3} \mathbf{Z}_{L_{3}}-\mathbf{I}_{1} \mathbf{Z}_{m_{13}}=0$$ $$\mathbf{I}_{1} \mathbf{Z}_{m_{13}}+\mathbf{I}_{3}\left(\mathbf{Z}_{3}+\mathbf{Z}_{L_{3}}\right)=0$$
In determinant form,
$$\begin{split} &\mathbf{I}_{1}\left(\mathbf{Z}_{1}+\mathbf{Z}_{L_{1}}\right) &-\mathbf{I}_{2} \mathbf{Z}_{m_{12}} &+\mathbf{I}_{3} \mathbf{Z}_{m_{13}} &=\mathbf{E}_{1} \\ -&\mathbf{I}_{1} \mathbf{Z}_{m_{12}} &+\mathbf{I}_{2} \left(\mathbf{Z}_{2}+\mathbf{Z}_{L_{12}}\right) &+ 0 &=0 \\ &\mathbf{I}_{1} \mathbf{Z}_{m_{13}} &+0 &+\mathbf{I}_{3}\left(\mathbf{Z}_{3}+\mathbf{Z}_{13}\right) &=0 \end{split}$$

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