To solve this circuit using mesh analysis, we need to define the mesh currents and set up equations based on Kirchhoff's Voltage Law (KVL) for each loop in the circuit. Here's how we can approach it step-by-step.

Step 1: Define Mesh Currents

Let's define three mesh currents:

1. $I_1$ in the loop on the left containing $V_1$.
2. $I_2$ in the middle loop containing $V_2$.
3. $I_3$ in the loop on the right containing $V_3$.

The direction of the 10A current source in the middle loop implies that $I_2 = 10 \text{ A}$.

Step 2: Set Up KVL Equations for Each Loop

Left Loop (Loop 1 with $V_1$)

Using KVL for the left loop, which contains the 10Ω and 5Ω resistors, we have: $V_1 = 10 \cdot I_1 + 5 \cdot (I_1 - I_2)$

Right Loop (Loop 3 with $V_3$)

For the right loop containing the 5Ω and 10Ω resistors: $V_3 = 10 \cdot (I_3 - I_2) + 5 \cdot I_3$

Middle Loop (Loop 2 with $V_2$)

Since $I_2 = 10 \, \text{A}$, we substitute this directly into the equations for the other loops.

Step 3: Solve the Equations

Let's substitute $I_2 = 10 \, \text{A}$ into the equations and solve for $I_1$ and $I_3$.

1. Equation for $V_1$: $V_1 = 10 \cdot I_1 + 5 \cdot (I_1 - 10)$

2. Equation for $V_3$: $V_3 = 10 \cdot (I_3 - 10) + 5 \cdot I_3$

Step 4: Substitute and Simplify

Let's simplify each equation to express $V_1$ and $V_3$ in terms of $I_1$ and $I_3$.

1. For $V_1$: $V_1 = 10I_1 + 5I_1 - 50 = 15I_1 - 50$

2. For $V_3$: $V_3 = 10I_3 - 100 + 5I_3 = 15I_3 - 100$

We would need additional information to determine $I_1$ and $I_3$ if they were unknown, such as voltage values or boundary conditions.

Would you like more details on any specific part of this solution? Here are a few questions that may expand on this:

1. How is Kirchhoff's Voltage Law applied in mesh analysis?
2. Can mesh analysis be used for circuits with dependent sources?
3. What if there were additional loops—how would that change the approach?
4. How does mesh analysis differ from nodal analysis?
5. Could this circuit be solved using Thevenin’s theorem instead?

Tip: When dealing with current sources within loops, directly assign the mesh current to the value of the current source for simplicity in calculations.