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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.

For the circuit shown in Fig.4, use the superposition theorem to determine the current I1.

Learn how to use the superposition theorem to calculate the current I1 in a circuit with two voltage sources and a current source. This problem covers circuit analysis concepts such as Ohm's Law and resistor combinations. Suitable for Grades 11-12.

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