To solve this problem using the superposition theorem, we need to analyze the circuit by considering each source independently and then summing the effects of each source. Here's the step-by-step solution:

Step 1: Circuit Analysis with Superposition Theorem

1. Turn off E2 (6V source): Replace E2 with a short circuit.

   • This leaves a 12V source (E1) in series with the 2Ω resistor ($R_1$) and a parallel combination with the 4Ω resistor ($R_2$) and the current source (3A).

2. Find the current $I_{1}'$ due to E1 (12V) alone:

   • Calculate the total resistance seen by E1 when E2 is shorted.
   • The 2Ω resistor ($R_1$) is in series with the parallel combination of 4Ω ($R_2$) and the 3A current source.
   • Parallel combination resistance: $\text{Since the current source is ideal, it sets the current through its path. So, for voltage calculation, only the 4Ω resistor affects the voltage.}$
   • Calculate $I_{1}'$ using Ohm's law.

3. Turn off E1 (12V source): Replace E1 with a short circuit.

   • This leaves the 6V source (E2) in series with $R_1$ (2Ω) and a parallel combination of $R_2$ (4Ω) and the current source (3A).

4. Find the current $I_{1}''$ due to E2 (6V) alone:

   • Again, determine the equivalent resistance in this setup and use Ohm's law to find the current contribution through $R_1$.

5. Superimpose the Results:

   • Combine the currents: $I_1 = I_{1}' + I_{1}''$

Step 2: Calculate Each Component

Let’s go through the calculations for each scenario:

1. With E1 active (12V):

   • $R_{\text{eq}} = 2 \Omega + \text{Equivalent resistance of parallel path}$
   • Find $I_{1}'$.

2. With E2 active (6V):

   • $R_{\text{eq}} = 2 \Omega + \text{Equivalent resistance of parallel path}$
   • Find $I_{1}''$.

Step 3: Solution Details

Let's solve for the values numerically. I'll do the math and provide the calculations.The current $I_1$ through the 2Ω resistor is determined by superimposing the effects of both sources:

1. Current contribution due to the 12V source ($I_1'$) is $6.0 \, \text{A}$.
2. Current contribution due to the 6V source ($I_1''$) is $3.0 \, \text{A}$.

By adding these contributions: $I_1 = I_1' + I_1'' = 6.0 \, \text{A} + 3.0 \, \text{A} = 9.0 \, \text{A}$

So, the total current $I_1$ through $R_1$ is $9.0 \, \text{A}$.

Would you like more details on any part of this process? Here are some related questions to explore further:

1. How does the superposition theorem simplify circuit analysis?
2. Why does an ideal current source impact the voltage distribution in a parallel path?
3. What happens if the current source value changes in this circuit?
4. How would you analyze this circuit if $R_1$ and $R_2$ were swapped?
5. What is the power dissipated in $R_1$ due to both sources?

Tip: The superposition theorem is particularly useful when multiple sources are present, as it allows you to analyze the effect of each source separately, simplifying the overall solution process.