Analyze the circuit shown in the image, identify the components and their values, and determine the currents I1, I2, and I3.

Let the top horizontal line be Node A with potential $V_A$.
Let the bottom horizontal line be Node B with potential $V_B$.
We set Node B as the reference node, so $V_B = 0V$.

Based on the current arrows in the diagram:

• Current $I_3$ enters Node B.
• Current $I_1$ leaves Node B.
• Current $I_2$ leaves Node B. Therefore, KCL at Node B states that the sum of currents entering the node equals the sum of currents leaving the node:

We will write expressions for $I_1$, $I_2$, and $I_3$ using Ohm's Law and considering the voltage sources in each branch. The current direction is from higher potential to lower potential. Since the arrows for $I_1, I_2, I_3$ point from Node A towards Node B (or through components to Node B), we assume $V_A$ is higher than $V_B$.

For the leftmost branch (containing the $2\Omega$ resistor, $15V$ source, and $3\Omega$ resistor):
Current $I_3$ flows from Node A, through the $2\Omega$ resistor, then through the $15V$ source (positive terminal up, negative down), then through the $3\Omega$ resistor, and finally to Node B.
The voltage drop across the resistors and the source in the direction of $I_3$ is:

Since $V_B = 0V$:

For the middle branch (containing the $4\Omega$ resistor and $10V$ source):
Current $I_2$ flows from Node A, through the $4\Omega$ resistor, then through the $10V$ source (positive terminal up, negative down), and finally to Node B.

Since $V_B = 0V$:

For the rightmost branch (containing the $5\Omega$ resistor and $5V$ source):
Current $I_1$ flows from Node A, through the $5\Omega$ resistor, then through the $5V$ source (positive terminal up, negative down), and finally to Node B.

Since $V_B = 0V$:

Substitute Equations 2, 3, and 4 into Equation 1:

To eliminate the denominators, multiply the entire equation by the least common multiple of 5 and 4, which is 20:

Rearrange the terms to solve for $V_A$:

Now substitute $V_A = 2V$ back into Equations 2, 3, and 4:
For $I_3$:

For $I_2$:

For $I_1$:

The negative signs indicate that the actual direction of current flow is opposite to the direction indicated by the arrows in the diagram.

The final answers are: