Question 5.21: Two magnetically coupled coils are connected in series and t......
Two magnetically coupled coils are connected in series and their total effective inductance is found to be 4.4 mH. When one coil is reversed in connection, the combined inductance drops to 1.6 mH. Here all the flux due to the first coil links the second coil but only 40% of the flux due to the second coil links with the first coil. Find the self-inductance of each coil and the mutual inductance between the coils.
We know that,
Equivalent inductance in series aiding, L _{ eq }= L _1+ L _2+2 M
Equivalent inductance in series opposing, L _{ eq }= L _1+ L _2-2 M
In series aiding connection, the equivalent inductance will be more than that in series opposing connection. Therefore, 4.4 mH is the equivalent inductance in series aiding and 1.6 mH is the equivalent inductance in series opposing.
\therefore L _1+ L _2+2 M =4.4 ……….(1)
L_1+L_2-2 M=1.6 ………(2)
Let us subtract equation (2) from (1).
\begin{aligned}& L _1+ L _2+2 M =4.4 \\ & L_1+L_2-2 M=1.6 \\& \begin{array}{rr}(-) & (-) \quad(+) \quad(-) \\\hline & 4 M =2.8 \end{array} \\& \therefore M =\frac{2.8}{4}=0.7 mH \\&\end{aligned} …..(3)
Here all the flux produced by coil-1 links coil-2.
∴ Coefficient of coupling between coils 1 and 2, k _1=\frac{\phi_{12}}{\phi_1}=1
Here only 40% of the flux produced by coil-2 links coil-1.
∴ Coefficient of coupling between coils 2 and , k _2=\frac{\phi_{21}}{\phi_2}=0.4
Now coefficient of coupling, k =\sqrt{ k _1 k _2}=\sqrt{1 \times 0.4}=\sqrt{0.4}
We know that,
M=k \sqrt{L_1 L_2} \Rightarrow M^2=k^2 L_1 L_2\therefore L _2=\frac{ M ^2}{ k ^2 L _1}=\frac{0.7^2}{(\sqrt{0.4})^2 L _1}=\frac{1.225}{ L _1} …..(4)
On substituting for M and L_2 from equations (3) and (4) in equation (1), we get,
L_1+\frac{1.225}{L_1}+(2 \times 0.7)=4.4
On multiplying the above equation by L_1 , we get,
\begin{aligned} & L _1^2+1.225+1.4 L _1=4.4 L _1 \Rightarrow L _1^2+1.4 L _1-4.4 L _1+1.225=0 \\ & \therefore \quad L _1^2-3 L _1+1.225=0 \end{aligned}
The above equation is a quadratic function of L_1. The roots of the quadratic equation will give the values of L_1.
The roots of the quadratic equation are,
\begin{aligned}L _1 & =\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1.225}}{2}=\frac{3 \pm 2.0248}{2} \\& =2.5124 \text { or } 0.4876 \\\therefore L _1 & =2.5124 mH \text { or } L _1=0.4876 mH \end{aligned}
From equation (1), we get
L _2=4.4- L _1-2 M
\text { When } L_1=2.5124 m H, L_2=4.4-2.5124-2 \times 0.7=0.4876 mH
\text { When } L _1=0.4876 mH , L _2=4.4-0.4876-2 \times 0.7=2.5124 mH
RESULT
Mutual inductance, M = 0.7 mH
Self-inductance of coil-1, L_1 = 2.5124 mH
Self-inductance of coil-2, L_2 = 0.4876 mH