Find I_(0) in Fig. 10.8 using mesh-analysis. Answer: 5.969/55.45 A

## Step 1: Identify the Meshes<br />### The circuit has two meshes. Let's label them as Mesh 1 and Mesh 2.<br />- **Mesh 1** includes the 8-ohm resistor, -j2 ohm reactance, and the current source \( 10 / 0^\circ \) A.<br />- **Mesh 2** includes the 6-ohm resistor, j4 ohm reactance, and the voltage source \( 50 / 30^\circ \) V.<br /><br />## Step 2: Write the Mesh Equations<br />### For Mesh 1:<br />The voltage around Mesh 1 can be written using Kirchhoff's Voltage Law (KVL):<br />\[ 8I_1 - j2I_1 + 10 = 0 \]<br />Simplifying, we get:<br />\[ (8 - j2)I_1 = -10 \]<br /><br />### For Mesh 2:<br />The voltage around Mesh 2 can be written using KVL:<br />\[ 6I_2 + j4I_2 - 50\angle30^\circ = 0 \]<br />Simplifying, we get:<br />\[ (6 + j4)I_2 = 50\angle30^\circ \]<br /><br />## Step 3: Solve the Mesh Equations<br />### Solving for \( I_1 \):<br />\[ I_1 = \frac{-10}{8 - j2} \]<br />Converting to polar form:<br />\[ 8 - j2 = \sqrt{8^2 + (-2)^2}\angle\tan^{-1}\left(\frac{-2}{8}\right) = \sqrt{68}\angle-\tan^{-1}\left(\frac{1}{4}\right) \]<br />\[ I_1 = \frac{-10}{\sqrt{68}\angle-\tan^{-1}\left(\frac{1}{4}\right)} \]<br />\[ I_1 = \frac{-10}{\sqrt{68}}\angle\tan^{-1}\left(\frac{1}{4}\right) \]<br />\[ I_1 = -\frac{10}{\sqrt{68}}\angle\tan^{-1}\left(\frac{1}{4}\right) \]<br /><br />### Solving for \( I_2 \):<br />\[ I_2 = \frac{50\angle30^\circ}{6 + j4} \]<br />Converting to polar form:<br />\[ 6 + j4 = \sqrt{6^2 + 4^2}\angle\tan^{-1}\left(\frac{4}{6}\right) = \sqrt{52}\angle\tan^{-1}\left(\frac{2}{3}\right) \]<br />\[ I_2 = \frac{50\angle30^\circ}{\sqrt{52}\angle\tan^{-1}\left(\frac{2}{3}\right)} \]<br />\[ I_2 = \frac{50}{\sqrt{52}}\angle(30^\circ - \tan^{-1}\left(\frac{2}{3}\right)) \]<br /><br />## Step 4: Calculate \( I_0 \)<br />### Since \( I_0 \) is the current through the series combination of the 8-ohm resistor and -j2 ohm reactance, it is equal to \( I_1 \).<br /><br />## Step 5: Convert Results to Rectangular Form<br />### Converting \( I_1 \) back to rectangular form:<br />\[ I_1 = -\frac{10}{\sqrt{68}}\cos(\tan^{-1}\left(\frac{1}{4}\right)) - j\frac{10}{\sqrt{68}}\sin(\tan^{-1}\left(\frac{1}{4}\right)) \]<br /><br />### Simplify the expression:<br />\[ I_1 = -1.212 - j0.303 \]