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To solve this circuit using the **superposition principle**, we need to analyze the effect of each independent source (the 40V voltage source and the 3A current source) separately while "turning off" the other sources. Turning off means:

- **For voltage sources**: Replace with a short circuit (since the voltage becomes zero).
- **For current sources**: Replace with an open circuit (since the current becomes zero).

We'll then sum the contributions from each source to find \( V_x \).

### Step 1: Contribution of the 40V Voltage Source
We turn off the 3A current source (replace it with an open circuit).

#### Circuit Simplification:
- The 3A current source is open, so we only consider the 40V voltage source with the resistors.
- There are two 10Ω resistors in parallel with each other, and this combination is in series with the 8Ω resistor.
  
#### Find Equivalent Resistance:
The two 10Ω resistors are in parallel:
\[
R_{\text{parallel}} = \frac{10 \times 10}{10 + 10} = 5\ \Omega
\]
This is in series with the 8Ω resistor:
\[
R_{\text{total}} = 8 + 5 = 13\ \Omega
\]

#### Find Current \( I \):
The current in the circuit due to the 40V voltage source is:
\[
I = \frac{V}{R_{\text{total}}} = \frac{40}{13}\ \text{A} \approx 3.08\ \text{A}
\]

#### Voltage Across 8Ω Resistor:
The voltage across the 8Ω resistor (which is \( V_x \) for this case) is:
\[
V_x = I \times 8 = \frac{40}{13} \times 8 = \frac{320}{13}\ \text{V} \approx 24.62\ \text{V}
\]

### Step 2: Contribution of the 3A Current Source
Now, we turn off the 40V voltage source (replace it with a short circuit).

#### Circuit Simplification:
- The 40V source is shorted, so the 8Ω resistor is in parallel with the two 10Ω resistors. 
- The current source will distribute between these paths.

#### Current Division:
The total resistance in the path with the 8Ω resistor and two 10Ω resistors in parallel can be used to divide the current from the current source.

- The two 10Ω resistors are still in parallel, giving an equivalent resistance of 5Ω as before.
- The total equivalent resistance is the parallel combination of 8Ω and 5Ω:
\[
R_{\text{eq}} = \frac{8 \times 5}{8 + 5} = \frac{40}{13}\ \Omega
\]

#### Current Through 8Ω Resistor:
Using current division to find the current through the 8Ω resistor:
\[
I_{8\Omega} = 3A \times \frac{5}{8 + 5} = 3A \times \frac{5}{13} = \frac{15}{13}\ \text{A} \approx 1.15\ \text{A}
\]

#### Voltage Across 8Ω Resistor:
The voltage across the 8Ω resistor due to the 3A current source is:
\[
V_x = I_{8\Omega} \times 8 = \frac{15}{13} \times 8 = \frac{120}{13}\ \text{V} \approx 9.23\ \text{V}
\]

### Step 3: Total Voltage \( V_x \)
Finally, we sum the contributions from both sources:
\[
V_x = 24.62\ \text{V} + 9.23\ \text{V} = 33.85\ \text{V}
\]

### Final Answer:
\[
V_x \approx 33.85\ \text{V}
\]

Would you like a more detailed explanation of any part? Here are some related questions:

1. How is the superposition principle used in other circuits?
2. What happens if we have dependent sources in the circuit?
3. Can we solve this circuit using mesh or nodal analysis for verification?
4. How do we calculate equivalent resistances in more complex circuits?
5. Why do we replace voltage sources with short circuits and current sources with open circuits?

**Tip:** When using the superposition principle, always calculate the effect of each source one at a time, and sum the contributions at the end.

Learn how to find Vx in an electrical circuit using the superposition theorem. This solution involves calculating the contribution of each source separately, simplifying parallel and series resistances, and applying Ohm's Law. Detailed step-by-step calculations provided.

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