Question:

rewrite the same as given below : Here's the neatly rewritten version of your handwritten solution using the **Mesh Analysis Method**: --- ### **Mesh Analysis Method** --- #### **For Loop Current $i_1$** Using mesh analysis, we get: $$ (1 - j2)i_1 - i_2 = 12\angle 0^\circ \tag{1} $$ --- #### **From the Circuit:** $$ v_{in} = 14\angle 0^\circ = j5 \cdot x_{i_3} + j2 \cdot x_{i_1} $$ And since: $$ i_g = i_2 = 2 \cdot v_{in} = 2 \left[14\angle 0^\circ + j5 \cdot x_{i_3} + j2 \cdot x_{i_1} \right] $$ $$ = 28\angle 0^\circ + j10 \cdot x_{i_3} + j4 \cdot x_{i_1} \tag{2} $$ --- #### **For Loop Currents $i_2$ and $i_3$:** Applying **supermesh analysis**, we get: $$ j3 \cdot x_{i_2} + \left[ (1 \angle 0^\circ)(i_2 - i_1) \right] - j5 \cdot x_{i_3} + 4(i_3 - i_4) = 0 $$ Simplifying: $$ - i_1 + (1 + j3)i_2 + (4 - j5)i_3 - 4i_4 = 0 \tag{3} $$ --- Here is the neatly rewritten version of your solution (page 2): --- ### **From the circuit,** $$ I_x = i_4 - i_5 = i_4 + 2\angle 0^\circ \quad \text{(since } i_5 = -2\angle 0^\circ \text{ from circuit)} $$ --- ### **For Loop Current $i_4$:** Using mesh analysis, we get: $$ j6 \cdot i_4 + 4(i_4 - i_3) + 3(i_4 + 2\angle 0^\circ) = 0 $$ Expanding: $$ -4i_3 + (7 + j6)i_4 = -6\angle 0^\circ \tag{4} $$ --- ### **From Equations (1), (2), (3), and (4):** Solving the system, we get: $$ i_1 = [2.27 + j\,4.935]\, \text{A} $$ $$ i_2 = [0.1502 + j\,0.338]\, \text{A} $$ $$ i_3 = [-0.8566 + j\,0.926]\, \text{A} $$ $$ i_4 = [-0.514 + j\,0.97]\, \text{A} $$ --- ### **Compute $I_x$:** $$ I_x = i_4 + 2\angle 0^\circ = [1.486 + j\,0.97]\, \text{A} $$ Convert to polar form: $$ I_x = 1.774 \angle 33.13^\circ \, \text{A} $$ --- ### **Compute $V_x$:** $$ V_x = 0.562 \angle 152.79^\circ \, \text{Volts} $$ ---