https://file.aiquickdraw.com/mathbot/file/1725975774646g6o2vjyn.jpg

To solve for the current \( I_{13} \) (the current through the 13 Ω resistor) using the **Current Division Principle (CDP)**, let's break down the circuit step by step. 

### Step 1: Analyze the circuit
From the image, the circuit consists of a 120V power source in series with a 5Ω resistor, which then splits into three parallel branches:

1. **Branch 1**: 7Ω resistor in series with an 8Ω resistor.
2. **Branch 2**: A single 10Ω resistor.
3. **Branch 3**: A 13Ω resistor in series with two parallel resistors: 11Ω and 12Ω, followed by a 14Ω resistor and a 15Ω resistor.

Our goal is to use CDP to find the current through the 13Ω resistor.

### Step 2: Simplify the circuit

1. **Simplify the third branch (containing the 13Ω resistor)**:
    - The two resistors (11Ω and 12Ω) are in parallel:
      \[
      R_{\text{parallel}} = \left( \frac{1}{11} + \frac{1}{12} \right)^{-1} = 5.739\ \Omega
      \]
    - Now, this equivalent resistance is in series with the 13Ω resistor:
      \[
      R_{\text{total (13Ω branch)}} = 13 + 5.739 = 18.739\ \Omega
      \]
    - This is then in series with 14Ω and 15Ω resistors:
      \[
      R_{\text{total (branch 3)}} = 18.739 + 14 + 15 = 47.739\ \Omega
      \]

2. **Simplify the other branches**:
    - The 7Ω and 8Ω resistors are in series:
      \[
      R_{\text{total (branch 1)}} = 7 + 8 = 15\ \Omega
      \]
    - The 10Ω resistor is standalone, so \( R_{\text{total (branch 2)}} = 10\ \Omega \).

### Step 3: Combine all the branches

Now, we have three resistances in parallel:
- \( R_{\text{branch 1}} = 15\ \Omega \)
- \( R_{\text{branch 2}} = 10\ \Omega \)
- \( R_{\text{branch 3}} = 47.739\ \Omega \)

The total equivalent resistance \( R_{\text{eq}} \) of the parallel combination is:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{15} + \frac{1}{10} + \frac{1}{47.739}
\]
Let's calculate \( R_{\text{eq}} \).It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can still try the following steps to complete the calculation:

1. Use the formula:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{15} + \frac{1}{10} + \frac{1}{47.739}
   \]
   to find the total equivalent resistance \(R_{\text{eq}}\) for the parallel combination.

2. Once you have \(R_{\text{eq}}\), use Ohm's Law to find the total current from the power source:
   \[
   I_{\text{total}} = \frac{120}{R_{\text{eq}}}
   \]
   (since the voltage is 120V).

3. Apply the **Current Division Principle** to find the current in the 13Ω branch:
   \[
   I_{13} = I_{\text{total}} \times \frac{R_{\text{other\ branches}}}{R_{\text{branch 3}}}
   \]

Do you have any questions about these steps? Here are some related questions to deepen your understanding:

1. What is the purpose of the Current Division Principle?
2. How does current divide in parallel circuits?
3. How can you simplify resistors in parallel?
4. What is the relationship between resistance and current in parallel circuits?
5. Can the voltage across different parallel branches vary in a circuit?

**Tip**: Always start by simplifying complex circuits step by step to make calculations easier.

Learn how to solve for the current in a 13Ω resistor using the Current Division Principle (CDP) in a circuit with parallel and series components. This solution covers key electrical concepts, including Ohm's Law and the formula for equivalent resistance. Suitable for advanced high school or college-level physics.

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