To determine $V_1$, $V_2$, and $V_3$ in this circuit using nodal analysis, let's label each node and apply Kirchhoff's Current Law (KCL) at each node where we need to find the voltage. The circuit has a current source of 10 A and resistors arranged as shown in the diagram.

Step-by-Step Solution

1. Assign Node Voltages:

   • Let $V_1$, $V_2$, and $V_3$ be the voltages at the respective nodes in the circuit as shown in the image.
   • We'll set the bottom of the circuit as the reference ground (0 V).

2. Apply KCL to Each Node:

Node $V_1$:

At node $V_1$: $\frac{V_1 - V_2}{5} + \frac{V_1}{10} = 0$ This equation accounts for the currents through the 5 Ω and 10 Ω resistors.

Node $V_2$:

At node $V_2$: $\frac{V_2 - V_1}{5} + \frac{V_2 - V_3}{10} = 10$ The current source contributes a current of 10 A leaving the node $V_2$.

Node $V_3$:

At node $V_3$: $\frac{V_3 - V_2}{10} + \frac{V_3}{5} = 0$ This equation represents the currents through the 10 Ω and 5 Ω resistors at node $V_3$.

Set Up the Equations:

We now have three equations from the KCL application:

1. $\frac{V_1 - V_2}{5} + \frac{V_1}{10} = 0$
2. $\frac{V_2 - V_1}{5} + \frac{V_2 - V_3}{10} = 10$
3. $\frac{V_3 - V_2}{10} + \frac{V_3}{5} = 0$

Solve the System of Equations:

Let’s simplify each equation:

1. For Node $V_1$: $2V_1 - V_2 = 0$ So, $V_2 = 2V_1$.

2. For Node $V_2$: Substitute $V_2 = 2V_1$ into this equation: $\frac{2V_1 - V_1}{5} + \frac{2V_1 - V_3}{10} = 10$ Simplifying, we get: $\frac{V_1}{5} + \frac{2V_1 - V_3}{10} = 10$

3. For Node $V_3$: Substitute