https://file.aiquickdraw.com/mathbot/file/1726479963102pd3cpti7.jpg

We can solve this problem using **mesh analysis**, where we will apply **Kirchhoff's Voltage Law (KVL)** to each loop.

Given:
- \(i_1\), \(i_2\), and \(i_3\) are the mesh currents for loops 1, 2, and 3 respectively.
- There's a dependent current source given as \(4i_0\), where \(i_0 = i_2 - i_3\) (as indicated in the circuit diagram).

### Step 1: Write KVL equations for each mesh.

#### Loop 1 (Mesh 1: containing \(i_1\))
For the first loop:
- The voltage drop across the 10Ω resistor: \(10(i_1 - i_2)\).
- The voltage drop across the 12Ω resistor: \(12i_1\).
- The supplied voltage: 24V.

Applying KVL to this loop:
\[
24 - 12i_1 - 10(i_1 - i_2) = 0
\]
Simplifying:
\[
24 - 12i_1 - 10i_1 + 10i_2 = 0
\]
\[
24 = 22i_1 - 10i_2 \quad \text{(Equation 1)}
\]

#### Loop 2 (Mesh 2: containing \(i_2\))
For the second loop:
- The voltage drop across the 10Ω resistor: \(10(i_2 - i_1)\).
- The voltage drop across the 4Ω resistor: \(4(i_2 - i_3)\).
- The voltage drop across the 24Ω resistor: \(24i_2\).

Applying KVL to this loop:
\[
10(i_2 - i_1) + 4(i_2 - i_3) + 24i_2 = 0
\]
Simplifying:
\[
10i_2 - 10i_1 + 4i_2 - 4i_3 + 24i_2 = 0
\]
\[
(10 + 4 + 24)i_2 - 10i_1 - 4i_3 = 0
\]
\[
38i_2 - 10i_1 - 4i_3 = 0 \quad \text{(Equation 2)}
\]

#### Loop 3 (Mesh 3: containing \(i_3\))
For the third loop, we have the dependent current source, which simplifies the analysis.

- The voltage drop across the 4Ω resistor: \(4(i_3 - i_2)\).
- The voltage of the dependent source is given as \(4i_0\), where \(i_0 = i_2 - i_3\), so the voltage is \(4(i_2 - i_3)\).

Applying KVL:
\[
4(i_3 - i_2) - 4(i_2 - i_3) = 0
\]
Simplifying:
\[
4i_3 - 4i_2 - 4i_2 + 4i_3 = 0
\]
\[
8i_3 - 8i_2 = 0
\]
\[
i_3 = i_2 \quad \text{(Equation 3)}
\]

### Step 2: Solve the system of equations

We now have three equations:
1. \(24 = 22i_1 - 10i_2\)
2. \(38i_2 - 10i_1 - 4i_3 = 0\)
3. \(i_3 = i_2\)

We can substitute \(i_3 = i_2\) into Equation 2 to reduce the system and solve for \(i_1\) and \(i_2\).

Let me calculate this for you.The loop currents are:
- \(i_1 = \frac{34}{27} \, \text{A} \approx 1.26 \, \text{A}\)
- \(i_2 = \frac{10}{27} \, \text{A} \approx 0.37 \, \text{A}\)
- \(i_3 = \frac{10}{27} \, \text{A} \approx 0.37 \, \text{A}\)

### Step 3: Dependent source calculation

The value of the dependent voltage source is \(4i_0\), where \(i_0 = i_2 - i_3\). Since \(i_2 = i_3\), the current \(i_0 = 0\), and hence the voltage of the dependent source is:
\[
4i_0 = 4 \times 0 = 0 \, \text{V}
\]

Thus, the dependent voltage source contributes **0 V** to the circuit.

Would you like further details on the solution? Here are some additional questions to expand on this:

1. How would the mesh equations change if the dependent source had a different multiplier?
2. Can mesh analysis be used with circuits that contain inductors or capacitors?
3. What is the impact of changing the 24Ω resistor on the mesh currents?
4. How would the circuit change if we replaced the dependent source with an independent voltage source?
5. How is mesh analysis different from nodal analysis?

**Tip**: Always check if dependent sources introduce additional constraints based on the relationship given for the controlling variable.

Find loop currents i1, i2, i3 and the value of dependent source using mesh analysis.

Learn how to solve for loop currents i1, i2, i3 and the value of a dependent voltage source using mesh analysis. This problem covers Kirchhoff's Voltage Law, Ohm's Law, and dependent source analysis, suitable for undergraduate electrical engineering students.

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