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Find the answers for I1, I2, and I3
But also, give me the solutions for I1-3' (Which is done via disabling the 7 A at the middle)
Solutions for I1-3" (which is done via disabling the battery volt at the left, and basing it on the middle volts)
Utilize the superposition theorem
I1 doesn't come across any resistors so its purely 7, but it goes inside the node, which results to it being a negative. I2 is leaving the node making it positive and passes through 2 and 1 ohms. I3 is also leaving the node, ensuring that it also passes through 3 and 1 ohms


To solve this circuit using the **superposition theorem**, we need to analyze it by considering each source independently, while all other sources are replaced with their internal resistances (which means disabling current sources by opening them, and voltage sources by shorting them).

### Step 1: Analyzing with only the 7V voltage source active (disabling the 7A current source)

1. **Replace the current source (7A) with an open circuit**.
2. Find \(I_1'\), \(I_2'\), and \(I_3'\) using only the 7V source.

#### Steps:
- **Apply Kirchhoff’s Voltage Law (KVL)** around the loops and nodes.
- Identify the loop currents based on the configuration and the given resistance values.
- Solve for the currents \(I_1'\), \(I_2'\), and \(I_3'\).

### Step 2: Analyzing with only the 7A current source active (disabling the 7V voltage source)

1. **Replace the 7V voltage source with a short circuit**.
2. Find \(I_1''\), \(I_2''\), and \(I_3''\) using only the 7A source.

#### Steps:
- Apply KVL and Kirchhoff’s Current Law (KCL) based on the new configuration.
- The resistors form different paths for the current, so we calculate the resulting currents when the voltage source is replaced by a wire.

### Step 3: Superposition theorem

Once we have \(I_1'\), \(I_2'\), \(I_3'\) from the first scenario (voltage source only) and \(I_1''\), \(I_2''\), \(I_3''\) from the second scenario (current source only), the total current for each loop is the algebraic sum of the individual currents:

\[
I_1 = I_1' + I_1''
\]
\[
I_2 = I_2' + I_2''
\]
\[
I_3 = I_3' + I_3''
\]

I will proceed to solve this system now by calculating these values. Let's begin with analyzing both conditions.Here are the partial solutions for the currents in both conditions:

### 1. **Voltage Source Active (7V only)**:
- The loop analysis yields the following relationships:
  - \( I_1' = 7 - 10 I_3' \)
  - \( I_2' = 5 I_3' \)

### 2. **Current Source Active (7A only)**:
- Kirchhoff's Current Law (KCL) at the node gives the sum of currents:
  - \( I_1'' + I_2'' + I_3'' = 7 \)

At this point, we still need to solve the system to find the exact values of the currents, which will give us the total current for \(I_1\), \(I_2\), and \(I_3\) using the superposition theorem.

I will now proceed to fully solve this system and determine the current values.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can walk you through the steps to solve it manually or provide further guidance based on the partial results we have so far. Let me know how you'd like to proceed!

Here are some questions you might explore further:
1. How does the superposition theorem work in more complex circuits?
2. Can the superposition theorem be applied to nonlinear components like diodes?
3. What are the limitations of using superposition for AC circuits?
4. How do voltage and current division techniques simplify circuit analysis?
5. What role do Kirchhoff’s laws play in understanding complex networks?

**Tip**: Always double-check for consistency in units and signs when solving electrical circuit problems to avoid simple mistakes!

Find the answers for I1, I2, and I3, as well as solutions for I1-3' (disabling the 7 A current source) and I1-3" (disabling the battery voltage). Utilize the superposition theorem.

Learn how to solve a circuit problem using the superposition theorem by finding I1, I2, and I3. This problem involves disabling the 7 A current source and the voltage source separately, using Kirchhoff's Laws for analysis. Suitable for college-level students.

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