Question:

organise the all content in organised manner Mesh Analysis Method 1. For Loop Current \(\ce{i_1}\)i_1 Applying mesh analysis: \(\ce{(1 - j2)i_1 - i_2 = 12\angle 0^\circ ......(1)}\)(1 - j2)i_1 - i_2 = 12\angle 0^\circ ......(1) 2. From the Circuit The input voltage is given by: \(\ce{v_{in} = 14\angle 0^\circ = j5 \cdot i_3 + j2 \cdot i_1}\)v_{in} = 14\angle 0^\circ = j5 \cdot i_3 + j2 \cdot i_1 Given that: \(\ce{i_g = i_2 = 2 \cdot v_{in} = 2 \left[14\angle 0^\circ + j5i_3 + j2i_1 \right]}\)i_g = i_2 = 2 \cdot v_{in} = 2 \left[14\angle 0^\circ + j5i_3 + j2i_1 \right] \(\ce{= 28\angle 0^\circ + j10i_3 + j4i_1 \tag{2}}\)= 28\angle 0^\circ + j10i_3 + j4i_1 \tag{2} 3. For Loop Currents \(\ce{i_2 and }\)i_2 and \(\mathrm{{i}_{{3}}}\)i_3 (Supermesh) Applying supermesh analysis: \(\ce{j3\,i_2 + \left[ 1\angle 0^\circ (i_2 - i_1) \right] - j5\,i_3 + 4(i_3 - i_4) = 0}\)j3\,i_2 + \left[ 1\angle 0^\circ (i_2 - i_1) \right] - j5\,i_3 + 4(i_3 - i_4) = 0 Simplifying: \(\mathrm{{i}_{{1}}+{\left({1}+{j}{3}\right)}{i}_{{2}}+{\left({4}-{j}{5}\right)}{i}_{{3}}-{4}{i}_{{4}}={0}\ldots\ldots\ldots\ldots\ldots\ldots{\left\lbrace{3}\right\rbrace}}\)i_1 + (1 + j3)i_2 + (4 - j5)i_3 - 4i_4 = 0 ..................{3} \(\ce{4. For Loop Current i_4}\)4. For Loop Current i_4From the circuit: \(\mathrm{{I}_{{x}}={i}_{{4}}-{i}_{{5}}={i}_{{4}}+{2}\angle{0}^{\circ}\quad\text{(since }\ {i}_{{5}}=-{2}\angle{0}^{\circ}\text{)}}\)I_x = i_4 - i_5 = i_4 + 2\angle 0^\circ \quad \text{(since } i_5 = -2\angle 0^\circ \text{)} \(\ce{Applying mesh analysis:}\)Applying mesh analysis: \(\mathrm{{j}{6},{i}_{{4}}+{4}{\left({i}_{{4}}-{i}_{{3}}\right)}+{3}{\left({i}_{{4}}+{2}\angle{0}^{\circ}\right)}={0}}\)j6\,i_4 + 4(i_4 - i_3) + 3(i_4 + 2\angle 0^\circ) = 0 \(\ce{Expanding:}\)Expanding: \(\mathrm{-{4}{i}_{{3}}+{\left({7}+{j}{6}\right)}{i}_{{4}}=-{6}\angle{0}^{\circ}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.{\left\lbrace{4}\right\rbrace}}\)-4i_3 + (7 + j6)i_4 = -6\angle 0^\circ ......................{4} 5. Solving the System (Equations 1–4) By solving the above equations, we obtain: \(\mathrm{{i}_{{1}}={2.27}+{j},{4.935}\ \text{A}}\)i_1 = 2.27 + j\,4.935 \ \text{A} \(\mathrm{{i}_{{2}}={0.1502}+{j},{0.338}\ \text{A}}\)i_2 = 0.1502 + j\,0.338 \ \text{A} \(\mathrm{{i}_{{3}}=-{0.8566}+{j},{0.926}\ \text{A}}\)i_3 = -0.8566 + j\,0.926 \ \text{A} \(\mathrm{{i}_{{4}}=-{0.514}+{j},{0.97}\ \text{A}}\)i_4 = -0.514 + j\,0.97 \ \text{A} 6. Finding \(\ce{I_x}\)I_x \(\mathrm{{I}_{{x}}={i}_{{4}}+{2}\angle{0}^{\circ}={1.486}+{j},{0.97}\ \text{A}}\)I_x = i_4 + 2\angle 0^\circ = 1.486 + j\,0.97 \ \text{A} \(\ce{In polar form:}\)In polar form: \(\mathrm{{I}_{{x}}={1.774}\angle{33.13}^{\circ}\ \text{A}}\)I_x = 1.774 \angle 33.13^\circ \ \text{A} 7. Finding \(\ce{V_x}\)V_x \(\mathrm{{V}_{{x}}={0.562}\angle{152.79}^{\circ}\ \text{Volts}}\)V_x = 0.562 \angle 152.79^\circ \ \text{Volts}