1) please use nodal approach to find out voltage of the 10 - ohm resistor (v_r) in the circuit. (20 points)

Explanation:

Step1: Select reference node

Choose the grounded node as the reference node. Let the voltage at the node across the 10 - ohm resistor be $V_R$.

Step2: Apply KCL at the node of $V_R$

The currents entering and leaving the node must sum to zero. The admittances of the resistors are used. The admittance of a resistor $R$ is $Y=\frac{1}{R}$. The current through the 10 - ohm resistor is $I_{10}=\frac{V_R}{10}$. The current from the current - source $3\angle - 30^{\circ}$ enters the node. For the other branches, we first find the equivalent impedance and then the current. The impedance of the series combination of $2\Omega$ and $3\Omega$ resistors is $Z_1 = 2 + 3=5\Omega$. The voltage across this combination is related to the other sources. The current from the $1\angle0^{\circ}$ current - source and the voltage - source $2\angle45^{\circ}$ also contribute to the KCL equation. The admittance of the $10\Omega$ resistor $Y_{10}=\frac{1}{10}S$. The KCL equation at the node of $V_R$ is: [3\angle - 30^{\circ}+\frac{V_R - 2\angle45^{\circ}}{5}= \frac{V_R}{10}] [3(\cos(- 30^{\circ})+j\sin(- 30^{\circ}))+\frac{V_R - 2(\cos45^{\circ}+j\sin45^{\circ})}{5}= \frac{V_R}{10}] [3\left(\frac{\sqrt{3}}{2}-j\frac{1}{2}\right)+\frac{V_R - 2\left(\frac{\sqrt{2}}{2}+j\frac{\sqrt{2}}{2}\right)}{5}= \frac{V_R}{10}] [ \frac{3\sqrt{3}}{2}-j\frac{3}{2}+\frac{V_R}{5}-\frac{2\sqrt{2}}{10}-j\frac{2\sqrt{2}}{10}=\frac{V_R}{10}] [ \frac{3\sqrt{3}}{2}-j\frac{3}{2}-\frac{\sqrt{2}}{5}-j\frac{\sqrt{2}}{5}=\frac{V_R}{10}-\frac{V_R}{5}] [ \frac{3\sqrt{3}}{2}-j\frac{3}{2}-\frac{\sqrt{2}}{5}-j\frac{\sqrt{2}}{5}=-\frac{V_R}{10}] [V_R = 10\left(\frac{\sqrt{2}}{5}+j\frac{\sqrt{2}}{5}+\frac{3}{2}j-\frac{3\sqrt{3}}{2}\right)] [V_R=\left(2\sqrt{2}-15\sqrt{3}\right)+j\left(2\sqrt{2} + 15\right)] [V_R\approx(2\times1.414 - 15\times1.732)+j(2\times1.414 + 15)] [V_R\approx(2.828-25.98)+j(2.828 + 15)] [V_R\approx - 23.152+j17.828] [|V_R|=\sqrt{(-23.152)^2+(17.828)^2}] [|V_R|=\sqrt{535.91+317.84}] [|V_R|=\sqrt{853.75}] [|V_R|\approx29.22] [ \theta=\tan^{-1}\left(\frac{17.828}{-23.152}\right)\approx142.4^{\circ}] [V_R\approx29.22\angle142.4^{\circ}]

Answer:

$V_R\approx29.22\angle142.4^{\circ}$