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To solve Question #1, where we are tasked with finding \( V_L \) using a process similar to the one in Example-60, we can apply the **superposition theorem**. This theorem allows us to analyze the circuit by considering one independent source at a time and then combining the results.

### Step-by-Step Process

#### 1. **Circuit Setup**

We are given two current sources and three resistors in the circuit:
- \( 5A \) current source connected to a \( 6\Omega \) resistor.
- \( 7A \) current source connected to a \( 2\Omega \) resistor.
- The \( 8\Omega \) resistor is where we want to find the voltage drop \( V_L \).

#### 2. **Apply Superposition Theorem**
   - **Deactivate the 7A source** (consider only the 5A source first).
   - **Deactivate the 5A source** (consider only the 7A source next).
   - Sum the individual contributions of the two cases to find the total \( V_L \).

### Case 1: Deactivate the 7A Source (Open Circuit the 7A Branch)
In this case, only the 5A source is active. The equivalent resistance seen by the 5A current source is:

\[
R_{\text{eq1}} = 6\Omega + 8\Omega = 14\Omega
\]

The current from the 5A source will flow through both the \( 6\Omega \) and \( 8\Omega \) resistors (series connection), and the voltage across the \( 8\Omega \) resistor is given by:

\[
V_{L1} = I \cdot R_{\text{8}\Omega} = 5A \times 8\Omega = 40V
\]

### Case 2: Deactivate the 5A Source (Open Circuit the 5A Branch)
Now consider only the 7A source. The equivalent resistance seen by the 7A current source is:

\[
R_{\text{eq2}} = 2\Omega + 8\Omega = 10\Omega
\]

The current from the 7A source will flow through both the \( 2\Omega \) and \( 8\Omega \) resistors (series connection), and the voltage across the \( 8\Omega \) resistor is:

\[
V_{L2} = I \cdot R_{\text{8}\Omega} = 7A \times 8\Omega = 56V
\]

### 3. **Combine Results from Superposition**
Since the two sources have opposing directions, we subtract the voltages:

\[
V_L = V_{L2} - V_{L1} = 56V - 40V = 16V
\]

### Final Answer:
The voltage across the \( 8\Omega \) resistor is \( V_L = 16V \).

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Would you like further details on any part of this solution, or do you have any questions?

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Here are five follow-up questions based on this problem:

1. How would the result change if the direction of one of the current sources was reversed?
2. Can we solve this circuit using Thevenin's Theorem instead of superposition?
3. What is the power dissipated by the \( 8\Omega \) resistor?
4. How do we handle circuits with dependent sources when applying superposition?
5. How would the voltage \( V_L \) change if the \( 8\Omega \) resistor was replaced with a \( 10\Omega \) resistor?

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**Tip:** When applying the superposition theorem, always deactivate independent sources by replacing voltage sources with short circuits and current sources with open circuits.

Solve for V_L using a process similar to Example-60.

Learn how to solve a circuit problem using the superposition theorem, applying it to find the voltage across an 8-ohm resistor with two current sources in parallel. A step-by-step solution demonstrates how to deactivate sources and calculate partial voltages, combining them to find V_L. Suitable for undergraduate electrical engineering students.

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