Question 4. Pod V, in the network shown in the given figure, where P.6240" V. Peste report your answer to the magnitude in partive and at angles are in the range of negative 180 days to postive 180 degrees ΤΩ 20 20 w V j40 140 -jla 2/0A w P . The value of V, 2

Q(4) Find v_{x} in the netwrok shown in the giver figure, (page 1 cohere p=4 \angle 40^{\circ} \mathrm{V},Solution:- By using source trans formation method. cursect source and Reactance across capacitor converted into voltage. Soure and series capacitive, reactance\begin{array}{l}I=2 \angle 0^{\circ} A \\x_{C}=-j i r \\\Rightarrow V=I X_{C} \\\Rightarrow V=210^{\circ} \times(-j 1) \\\Rightarrow V=-2 j \\\Rightarrow V=-2 \angle 90^{\circ} \\\Rightarrow V=(-1) \times 2\left\lfloor 90^{\circ}\right. \\8^{2}=-1 \\\Rightarrow \quad V=i^{2} \times 22^{\circ} \\\Rightarrow \quad V=\left(1 \angle 90^{\circ}\right)^{2} \times 2\left\lfloor 90^{\circ}\right. \\\end{array}\begin{array}{l}\Rightarrow V=\angle \angle 90^{\circ} \times 140^{\circ} \times 2 \angle 90^{\circ} \\\Rightarrow V=1 \angle 180^{\circ} \times 2 \angle 90^{\circ} \\\therefore V=2 \angle 270^{\circ} \\\Rightarrow V=2 \angle-360^{\circ}+270^{\circ} \\\therefore V=2 \angle-90^{\circ} \mathrm{V}\end{array}Page 2)That means we can say that rotationthat means we can say whet\angle 270^{\circ}=\angle-90^{\circ}Applying KVL in mesh (01)-4 \angle 40^{\circ}+2 I_{1}+j 4 I_{1}-j 1 I_{2}=0(2+j 4) I_{1}-j I_{2}=4 \angle 40^{\circ}page 3Applying KVL in mesh (02)\begin{array}{c}2 I_{2}-j I_{2}+2 L-90^{\circ}+j 4 I_{2}-j 1 I_{1}=0 \\-j+I_{1}+2 I_{2}-j I_{2}+j 4 I_{2}=-2 L-90^{\circ} \\-j 1 I_{1}+2 I_{2}+3 j I_{2}=-2 L-90^{\circ} \\-j 1 I_{1}+(2+3 j) I_{2}=-2 L-90^{\circ} \\-j 1 I_{1}+(2+3 j) I_{2}=(-1) \times 2 L-90^{\circ} \\-j 1 I_{1}+(2+3 j) I_{2}=\left(j^{2}\right) \times 2 L-90^{\circ} \\j^{2}=-1\end{array}\begin{array}{l}-j 1 I_{1}+(2+3 j) I_{2}=\left(1290^{\circ}\right)^{2} \times 2 \angle 90^{\circ} \\-j+I_{1}+(2+3 j) I_{2}=1 \angle 180^{\circ} \times 2 \angle-90^{\circ} \\-j \perp I_{1}+(2+3 j) I_{2}=2 \angle 180^{\circ}-90^{\circ} \\-j d I_{1}+(2+3 j) I_{2}=2 L 90^{\circ}\end{array}(ii)From equation (i) and equation (ii)Page 4\begin{array}{l}(2+j 4) I_{1}-j I_{2}=4\left\lfloor 40^{\circ}-\right.\text { polar form } \\(2+j 4) I_{1}-j I_{2}=4\left[\cos 40^{\circ}+j \sin 40^{\circ}\right] \\(2+j 4) I_{1}-j I_{2}=4[0.766+j 0.6427] \\(2+j 4) I_{1}-j I_{2}=4 \times 0.766+j 0.6427 \times 4\end{array}(2+j 4) I_{1}-j I_{2}=3.064+j 2.5708 \text { (Rectangular form) }so, (2+j 4) I_{1}-j I_{2}=(3.064+j 2.5708)-j I_{1}+(2+3 j) I_{2}=2 jsolving above two equation by cramer RuleAbove two equation writien as a matrix formA X=B\left[\begin{array}{cc}2+j 4 &amp; -j \\-j &amp; (2+3 j)\end{array}\right]\left[\begin{array}{l}I_{1} \\I_{2}\end{array}\right]=\left[\begin{array}{c}3.064+j 2.5708 \\2 j\end{array}\right]\Delta=\left|\begin{array}{cc}2+j 4 &amp; -j \\-j &amp; (2+3 j\end{array}\right|\begin{array}{l}\Delta=(2+4 j)(2+3 j)-(-j) \times(-j) \\ \Delta=(4+6 j+8 j+3 j \times 4 j)-\left(+j^{2}\right) \\ \Delta=(4+14 j-12)-(-1) \\ \Delta=(14 j-8)+1 \\ \Delta=14 j-8+1 \\ \Delta=-7+14 \mathrm{j} \rightarrow \text { (rectangular form) } \\ \Delta=\sqrt{(7)^{2}+(14)^{2}}-\tan ^{-1}\left(\frac{14}{7}\right)+180^{\circ} \\ \Delta=\sqrt{49+196}<-\tan ^{-1}(2)+180^{\circ} \\ \Delta=\sqrt{245}<180^{\circ}-\tan ^{-1} 2 \\ \Delta=15.6524<180^{\circ}-63.43^{\circ} \\ \Delta=15.6524<116.57^{\circ} \quad \text { (polar form) } \\ \Delta_{2}=\left|\begin{array}{cc}2+j 4 &amp; 3.064+j 2.5708 \\ -j &amp; 2 j\end{array}\right| \\\end{array}\begin{array}{l}\Delta_{1}=(2+j 4) \times 2 j+j(3.064+j 2.5708) \text { page } \\ \Delta_{2}=4 j-8+3.064 j-2.5708 \\ \Delta_{2}=-8-2.5708+4 j+3.064 j \\ \Delta_{2}=-10.5708+j 7.064 \text { [Rectargular form] } \\ \Delta_{2}=\sqrt{(10.5708)^{2}+(7.064)^{2}\left(-\tan ^{-1}\left(\frac{7.064}{10.5708}\right)+180^{\circ}\right.} \\ \Delta_{2}=\sqrt{(111.7418+49.90)}-\tan ^{-1}(0.6682)+180^{\circ} \\ \Delta_{2}=\sqrt{161.6418}-33 \cdot 75+180^{\circ} \\ \Delta_{2}=12.713842\left(180^{\circ}-33.75^{\circ}\right. \\ D_{2}=12.713842\left\lfloor 146.25^{\circ} \text { [polar form] }\right. \\ I_{2}=\frac{\Delta_{2}}{\Delta} \\ I_{2}=\frac{12.713842 \angle 146.25^{\circ}}{15.6524 \angle 116.57^{\circ}} \\ I_{2}=0.8122615 / 146.25^{\circ}-116.57^{\circ} \\ I_{2}=0.8122615 \angle 29.68^{\circ} \mathrm{A} \rightarrow \text { Approximate } \\\end{array}V_{x}=2 \times I_{2}v_{x}=2 \times\left(0.8122615 \angle 29.68^{\circ}\right)v_{x}=1.624523 \angle 29.68^{\circ} \mathrm{V} \rightarrow \text { Approximate }\begin{aligned}\text { Magnitude } &amp; =1.6245^{\circ} 3 \\\text { angle } &amp; =29.68^{\circ} \mathrm{V}\end{aligned}Always Remember\begin{aligned}a+i b &amp; =\sqrt{a^{2}+b^{2}}\left\lfloor\tan ^{-1}(b / a)\right. \\a-i b &amp; =\sqrt{a^{2}+b^{2}}\left[-\tan ^{-1}(b / a)\right. \\-a+i b &amp; =\sqrt{a^{2}+b^{2}}\left\lfloor-\tan ^{-1}(b / a)+180^{\circ}\right. \\-a-i b &amp; =\sqrt{a^{2}+b^{2}}\left\lfloor\tan ^{-1}(b / a)-180^{\circ}\right.\end{aligned}Final Auswer ...