The problem requires us to find the node voltages $V_1$, $V_2$, and $V_3$ in the given circuit using nodal analysis.

1. Problem Description

The problem requires us to find the node voltages $V_1$, $V_2$, and $V_3$ in the given circuit using nodal analysis.

2. Solution Steps

We will apply Kirchhoff's Current Law (KCL) at each node $V_1$, $V_2$, and $V_3$.

At node $V_1$:

The incoming current is 5A. The outgoing currents are $\frac{V_1 - V_2}{1}$ and $\frac{V_1 - V_3}{1}$.

Applying KCL:

$5 = \frac{V_1 - V_2}{1} + \frac{V_1 - V_3}{1}$

$5 = V_1 - V_2 + V_1 - V_3$

$2V_1 - V_2 - V_3 = 5$ (Equation 1)

At node $V_2$:

The incoming currents are $\frac{V_1 - V_2}{1}$ and $\frac{V_3 - V_2}{1}$. The outgoing current is $\frac{V_2}{4}$.

Applying KCL:

$\frac{V_1 - V_2}{1} + \frac{V_3 - V_2}{1} = \frac{V_2}{4}$

$V_1 - V_2 + V_3 - V_2 = \frac{V_2}{4}$

$V_1 - 2V_2 + V_3 = \frac{V_2}{4}$

$V_1 - \frac{9}{4}V_2 + V_3 = 0$

$4V_1 - 9V_2 + 4V_3 = 0$ (Equation 2)

At node $V_3$:

The incoming current is $\frac{V_1 - V_3}{1}$ and $\frac{V_2 - V_3}{1}$. The outgoing current is 10A.

Applying KCL:

$\frac{V_1 - V_3}{1} + \frac{V_2 - V_3}{1} = 10$

$V_1 - V_3 + V_2 - V_3 = 10$

$V_1 + V_2 - 2V_3 = 10$ (Equation 3)

Now we have three equations with three unknowns:

Equation 1: $2V_1 - V_2 - V_3 = 5$

Equation 2: $4V_1 - 9V_2 + 4V_3 = 0$

Equation 3: $V_1 + V_2 - 2V_3 = 10$

Multiply Equation 1 by 4:

$8V_1 - 4V_2 - 4V_3 = 20$ (Equation 4)

Add Equation 2 and Equation 4:

$12V_1 - 13V_2 = 20$ (Equation 5)

Multiply Equation 3 by 4:

$4V_1 + 4V_2 - 8V_3 = 40$ (Equation 6)

Subtract Equation 2 from Equation 6:

$-13V_2 - 12V_3 = 40$

$13V_2 + 12V_3 = -40$ (Equation 7)

From Equation 1, $V_3 = 2V_1 - V_2 - 5$.

Substitute this into Equation 3:

$V_1 + V_2 - 2(2V_1 - V_2 - 5) = 10$

$V_1 + V_2 - 4V_1 + 2V_2 + 10 = 10$

$-3V_1 + 3V_2 = 0$

$V_1 = V_2$

Substitute $V_1 = V_2$ in equation 5:

$12V_2 - 13V_2 = 20$

$-V_2 = 20$

$V_2 = -20$

$V_1 = -20$

Now, substitute $V_1 = -20$ and $V_2 = -20$ into Equation 3:

$-20 - 20 - 2V_3 = 10$

$-40 - 2V_3 = 10$

$-2V_3 = 50$

$V_3 = -25$

3. Final Answer

$V_1 = -20 V$

$V_2 = -20 V$

$V_3 = -25 V$