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.