Find Voltage across $ 5 \Omega $ resistor using mesh Analysis.

Solution:

$ \begin{aligned} \\ &X_M=K \sqrt{X_1 \cdot X_2}=K \sqrt{(5)(10)}\\\\ &\therefore=0.8 \sqrt{50}\\\\ &=5.65 \Omega\\\\ &\text { KVL to loop (1) }\\\\ &50-j 5 I_1+j 5.65 I_2-2\left(I_1-I_2\right)+j 4\left(I_1-I_2\right)\\\\ &-(2+j) I_1+(2+j 1.65) I_2=-50...(1)\\\\ &\text { KVL to loop (2) }\\\\ &-j 10 I_2+j 5.65 I_1-5 I_2+j 4\left(I_2-I_1\right)-2\left(I_2-I_1\right).\\\\ &(2+j 1.65) I_1-(7+j 6) I_2=0....(2)\\\\ &\text { Using Cramer's rule of determinant }\\ \end{aligned}\\ $

$ \begin{aligned} \\ &\Delta=\left|\begin{array}{ll} (2+j) & (2+j 1.65) \\\\ (2+j 1.65) & -(7+j 6) \\ \end{array}\right| \\\\ &=(2+j)(7+j 6)-(2+j 1.65)^2 \\\\ &=14.105 \lfloor 61.53 \\ \end{aligned} \\ $

$ \begin{aligned} \\ &\Delta_2=\left|\begin{array}{cc} -(2+j) & -50 \\\\ (2+j 1.65) & 0 \\ \end{array}\right|\\\\ &=0+(50)(2+j 1.65)\\\\ &=129.638 \lfloor 9.52 \\\\ &\therefore I_2=\frac{\Delta_2}{\Delta}=\frac{129.6381 \lfloor 39 .52}{1 4 . 1 0 5 \lfloor { 6 1 . 5 3 }}\\\\ \end{aligned}\\ $

$I_2=9.193 \lfloor 22.01 \mathrm{Amp}$

voltage across $ 5 \Omega $ resistor is

$ \begin{aligned} \\ V_5 &=(5)\left(I_2\right) \\\\ &=(5)(9.193 \lfloor -22.01) \\\\ V_5 &=45.965 \lfloor -22.01 \text { Volts } \\ \end{aligned} $