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To solve the given circuit using the mesh analysis, we'll define and solve the system of equations based on the three mesh currents labeled \( I_1 \), \( I_2 \), and \( I_3 \).

### Step 1: Define the meshes and apply Kirchhoff’s Voltage Law (KVL)
The circuit consists of three meshes, and we will apply KVL to each one to form a system of equations. 

#### Mesh 1 (Current \( I_1 \)):
This mesh includes the 30V voltage source, the 2Ω resistor, and the 6Ω resistor (shared with Mesh 2). The KVL equation for Mesh 1 is:
\[
30V - 2\Omega(I_1) - 6\Omega(I_1 - I_2) = 0
\]
Simplifying this:
\[
30 = 2I_1 + 6(I_1 - I_2)
\]
\[
30 = 8I_1 - 6I_2 \tag{1}
\]

#### Mesh 2 (Current \( I_2 \)):
This mesh includes the 6Ω resistor (shared with Mesh 1), the 2Ω resistor, and the 6Ω resistor (shared with Mesh 3). The KVL equation for Mesh 2 is:
\[
-6\Omega(I_2 - I_1) - 2\Omega(I_2) - 6\Omega(I_2 - I_3) = 0
\]
Simplifying this:
\[
-6(I_2 - I_1) - 2I_2 - 6(I_2 - I_3) = 0
\]
\[
-6I_2 + 6I_1 - 2I_2 - 6I_2 + 6I_3 = 0
\]
\[
6I_1 - 14I_2 + 6I_3 = 0 \tag{2}
\]

#### Mesh 3 (Current \( I_3 \)):
This mesh includes the 6Ω resistor (shared with Mesh 2), the 4Ω resistor, and the 4Ω resistor. The KVL equation for Mesh 3 is:
\[
-6\Omega(I_3 - I_2) - 4\Omega(I_3) - 4\Omega(I_3) = 0
\]
Simplifying this:
\[
-6(I_3 - I_2) - 4I_3 - 4I_3 = 0
\]
\[
-6I_3 + 6I_2 - 8I_3 = 0
\]
\[
6I_2 - 14I_3 = 0 \tag{3}
\]

### Step 2: Solve the system of equations
Now we have the following system of equations:
\[
8I_1 - 6I_2 = 30 \tag{1}
\]
\[
6I_1 - 14I_2 + 6I_3 = 0 \tag{2}
\]
\[
6I_2 - 14I_3 = 0 \tag{3}
\]

Let's solve this system of equations. I'll calculate the values of \( I_1 \), \( I_2 \), and \( I_3 \).The solutions for the mesh currents are:

- \( I_1 = \frac{600}{97} \, \text{A} \approx 6.19 \, \text{A} \)
- \( I_2 = \frac{315}{97} \, \text{A} \approx 3.25 \, \text{A} \)
- \( I_3 = \frac{135}{97} \, \text{A} \approx 1.39 \, \text{A} \)

These are the values of the mesh currents in the circuit.

Would you like more details or have any questions? Here are some related questions to expand on the topic:

1. How can mesh analysis be applied to circuits with more than three meshes?
2. What is the difference between mesh analysis and nodal analysis?
3. How would the solution change if one of the resistors was replaced with a different value?
4. Can you apply mesh analysis to circuits with current sources?
5. What are the limitations of using mesh analysis in complex circuits?

**Tip**: When solving circuit problems using mesh analysis, always double-check the sign conventions for voltage drops and rises across shared components between meshes.

Determine Io using mesh analysis in the given circuit (Figure 3.38). The circuit consists of a 30V voltage source and resistors (2Ω, 4Ω, 6Ω) connected in a complex arrangement with three mesh currents: I1, I2, and I3.

Learn how to determine the current Io in a complex circuit using mesh analysis. This step-by-step guide covers Kirchhoff's Voltage Law and Ohm's Law to solve for the mesh currents I1, I2, and I3. Suitable for high school and college-level students.

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