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To find \( V_L \) in the given circuit using the superposition theorem, we need to consider the contribution of each independent source separately and then sum the effects.

### Step 1: Analyzing the circuit with the 2V voltage source only

1. **Turn off the 5A current source:** Replace it with an open circuit.
2. **Find the equivalent resistance seen by the 2V source:**

   - The 2Ω resistor is in series with the 1Ω resistor on the top branch.
   - The 1Ω resistor is in parallel with the series combination of 1Ω and 1Ω resistors on the right branch.

   \[
   R_{eq1} = 2Ω + 1Ω = 3Ω
   \]
   The right branch has:
   \[
   R_{right} = 1Ω + (1Ω \parallel 1Ω) = 1Ω + \left(\frac{1Ω \cdot 1Ω}{1Ω + 1Ω}\right) = 1Ω + 0.5Ω = 1.5Ω
   \]
   \[
   R_{total} = 3Ω \parallel 1.5Ω = \frac{3Ω \cdot 1.5Ω}{3Ω + 1.5Ω} = \frac{4.5Ω}{4.5Ω} = 1Ω
   \]

3. **Find the current \( I_{1} \) from the 2V source:**

   \[
   I_{1} = \frac{2V}{1Ω} = 2A
   \]

4. **Determine the voltage drop across the 1Ω resistor on the right branch:** 

   Since the resistors are in parallel, the voltage drop across each resistor is the same as the source:
   \[
   V_{1Ω(right)} = 2V
   \]

### Step 2: Analyzing the circuit with the 5A current source only

1. **Turn off the 2V voltage source:** Replace it with a short circuit.
2. **Find the equivalent resistance seen by the 5A current source:**

   - The left branch has 2Ω in series with the parallel combination of the 1Ω and 1Ω resistors.
   - The right branch has a single 1Ω resistor.

   \[
   R_{left} = 2Ω + (1Ω \parallel 1Ω) = 2Ω + 0.5Ω = 2.5Ω
   \]

3. **Determine the voltage across the left branch:**

   Since the total resistance is the sum of the left and right branches in parallel,
   \[
   R_{total} = 2.5Ω \parallel 1Ω = \frac{2.5Ω \cdot 1Ω}{2.5Ω + 1Ω} = \frac{2.5Ω}{3.5Ω} = 0.714Ω
   \]
   Using Ohm's Law for the total current,
   \[
   V_{total} = I_{source} \cdot R_{total} = 5A \cdot 0.714Ω = 3.57V
   \]

   The voltage across each parallel branch is the same.

### Step 3: Combining the results using superposition

1. **Voltage \( V_{L} \) due to the 2V source:** 2V (as found earlier).
2. **Voltage \( V_{L} \) due to the 5A current source:** 3.57V (as found above).

**Combining both:**

\[
V_{L} = V_{L(2V)} + V_{L(5A)} = 2V + 3.57V = 5.57V
\]

Thus, the voltage \( V_L \) in the circuit using the superposition theorem is \( 5.57V \).

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Would you like more details or have any questions about the steps? Here are some related questions that might interest you:

1. What is the principle behind the superposition theorem?
2. How can we simplify a circuit with both series and parallel resistors?
3. How do you calculate the equivalent resistance of parallel resistors?
4. What happens to a voltage source when it is "turned off" in superposition analysis?
5. What happens to a current source when it is "turned off" in superposition analysis?
6. How do you calculate the total current in a circuit with multiple branches?
7. Why is the superposition theorem useful in circuit analysis?
8. Can the superposition theorem be used for non-linear components?

**Tip:** Always double-check the configuration of your circuit after "turning off" each independent source to ensure correct analysis.

Learn how to find VL in a circuit using the superposition theorem. Detailed steps involve analyzing series and parallel circuits, applying Ohm's Law, and combining effects from independent voltage and current sources. Suitable for undergraduate-level students and electronics enthusiasts.

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