Question 5.27: Find the mutual reactance Xm in the coupled coils shown in F......
Find the mutual reactance \text{X}_m in the coupled coils shown in Fig. 1, if the average power in 8 Ω resistance is 100 W.
Let us assume two mesh currents \bar{I}_1 \text { and } \bar{I}_2 as shown in Fig. 2. Now the current enters at the dotted end in one coil and leaves at the dotted end in the other coil. So the fluxes are opposing. Also, the mesh currents are in the same orientation. Hence, it is group-1 coupled coil. The electrical equivalent of the coupled circuit is shown in Fig. 3.
With reference to Fig. 3, the mesh basis matrix equation is,
\left[\begin{array}{rr} 5+j 5-j X_m+j X_m & -j X_m \\ -j X_m & j X_m+j 12-j X_m+8 \end{array}\right]\left[\begin{array}{r} \bar{I}_1 \\ \bar{I}_2 \end{array}\right]=\left[\begin{array}{r} 100 \angle 0^{\circ} \\ 0 \end{array}\right]
\left[\begin{array}{cr} 5+ j 5 & – j X _{ m } \\ – j X _{ m } & 8+ j 12 \end{array}\right]\left[\begin{array}{l} \overline{ I }_1 \\ \overline{ I }_2 \end{array}\right]=\left[\begin{array}{r} 100 \\ 0 \end{array}\right]
\text { Now, } \Delta=\left|\begin{array}{ll} 5+j 5 & -j X_m \\ -j X_m & 8+j 12 \end{array}\right|=(5+j 5) \times(8+j 12)-\left(-j X_m\right)^2
=-20+j 100+X_m^2=X_m^2-20+j 100
\Delta_2=\left|\begin{array}{lr} 5+j 5 & 100 \\ -j X_m & 0 \end{array}\right|=0-\left(-j X_m\right) \times 100=j 100 X_m
\text { Now, } \overline{ I }_2=\frac{\Delta_2}{\Delta}=\frac{ j 100 X _{ m }}{ X _{ m }^2-20+ j 100} …..(1)
Given that,
Power in 8 Ω resistance = 100 W
Here, Power in 8 Ω resistance =\left|\overline{ I }_2\right|^2 \times 8
\therefore\left|\bar{I}_2\right|^2 \times 8=100Using equation (1)
\left|\frac{j 100 X_m}{X_m^2-20+j 100}\right|^2 \times 8=100∴ \left(\frac{100 X_m}{\sqrt{\left(X_m^2-20\right)^2+100^2}}\right)^2 \times 8=100 \Rightarrow \frac{80000 X_m^2}{\left(X_m^2-20\right)^2+100^2}=100
\therefore\left( X _{ m }^2-20\right)^2+100^2=\frac{80000 X _{ m }^2}{100} \Rightarrow X _{ m }^4+400-40 X _{ m }^2+10000=800 X _{ m }^2
\therefore X _{ m }^4-40 X _{ m }^2-800 X _{ m }^2+10400=0 \Rightarrow X _{ m }^4-840 X _{ m }^2+10400=0
\text { Let, } X_m^2=X \text { Now, } X^2-840 X+10400=0The roots of the quadratic are,
\begin{aligned} X & =\frac{-(-840) \pm \sqrt{(-840)^2-4 \times 10400}}{2}=\frac{840 \pm 814.862}{2} \\ & =827.431 \text { or } 12.569 \end{aligned}
Let us take smaller value of X for realizability.
Now, \begin{aligned} & X_m^2=X \\ & \therefore X_m=\sqrt{X}=\sqrt{12.569}=3.5453 \Omega \end{aligned}
RESULT
\text { Mutual reactance, } X _{ m }=3.5453 \Omega