The image displays an electrical circuit with two voltage sources $E_1 = 8V$ and $E_2 = 24V$, along with three resistors: $R_1 = 2\Omega$, $R_2 = 1\Omega$, and $R_3 = 4\Omega$.

This is a parallel circuit, with the following components:

• $R_1$ and $R_2$ are directly connected to their respective voltage sources $E_1$ and $E_2$.
• $R_3$ shares a node between both branches.

To find the current through each resistor, we can apply Ohm's Law:

$I = \frac{V}{R}$

We calculate the current through each resistor as follows:

1. Current through $R_1$:

This resistor is in the branch with the 8V source:

$I_1 = \frac{E_1}{R_1} = \frac{8V}{2\Omega} = 4A$

2. Current through $R_2$:

This resistor is in the branch with the 24V source:

$I_2 = \frac{E_2}{R_2} = \frac{24V}{1\Omega} = 24A$

3. Current through $R_3$:

$R_3$ is shared between the two sources. To find the current through $R_3$, we need to analyze the potential difference across it, considering the voltages at the junction between the two sources.

If we assume the potential difference across $R_3$ is due to the difference between the two voltages, we get:

$\Delta V_{R_3} = E_2 - E_1 = 24V - 8V = 16V$

Then the current through $R_3$ is:

$I_3 = \frac{\Delta V_{R_3}}{R_3} = \frac{16V}{4\Omega} = 4A$

Summary of Currents:

• $I_1 = 4A$
• $I_2 = 24A$
• ( I_3 =