To solve for the loop currents $i_1$, $i_2$, $i_3$, and the dependent current source value using mesh analysis, we will proceed step by step:

1. Label the Mesh Currents:

• Let the loop currents be $i_1$, $i_2$, and $i_3$ as shown in the diagram.
• The dependent current source is given as $4i_0$, where $i_0$ is the current in the 4Ω resistor.

2. Mesh Analysis (KVL) Equations:

We'll write Kirchhoff’s Voltage Law (KVL) equations for each of the three loops.

Loop 1 (Mesh 1: $i_1$):

For the left-most loop with 24V source, $10 \Omega$ and $12 \Omega$ resistors: $24 - 10i_1 - 12(i_1 - i_3) = 0$ Simplifying: $24 - 22i_1 + 12i_3 = 0$ 22i_1 - 12i_3 = 24 \tag{1}

Loop 2 (Mesh 2: $i_2$):

For the middle loop, with the dependent source $4i_0$ and resistors $4 \Omega$ and $24 \Omega$: $24i_2 + 4(i_2 - i_3) = 0$ Simplifying: $28i_2 - 4i_3 = 0$ 7i_2 = i_3 \tag{2}

Loop 3 (Mesh 3: $i_3$):

For the right-most loop, with the dependent current source and resistors $12 \Omega$ and $4 \Omega$: [ 12(i_3 - i_1) + 4(i_3 - i