https://file.aiquickdraw.com/mathbot/file/1728930189550xvl5af6n.jpg

To solve this problem, we need to find the equivalent resistance for the circuit, starting from the left-most terminal. Let's break down the circuit step by step.

### Circuit Breakdown:
- **Resistor values given:**
  - \( R_1 = 7.2 \, \Omega \)
  - \( R_2 = 0.5 \, \Omega \)
  
1. **Simplify the parallel section:**
   - The parallel combination of the 3 resistors \( 2 \, \Omega \), \( 3 \, \Omega \), and \( 1 \, \Omega \) on the right side of the circuit needs to be computed first.

   The formula for the equivalent resistance \( R_{\text{eq}} \) for resistors in parallel is:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
   \]
   where \( R_1 = 2 \, \Omega \), \( R_2 = 3 \, \Omega \), and \( R_3 = 1 \, \Omega \).

   So,
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{1}
   \]
   \[
   \frac{1}{R_{\text{eq}}} = \frac{3}{6} + \frac{2}{6} + 1 = \frac{3}{6} + \frac{2}{6} + \frac{6}{6} = \frac{11}{6}
   \]
   Therefore, the equivalent resistance for this combination is:
   \[
   R_{\text{eq}} = \frac{6}{11} \, \Omega \approx 0.545 \, \Omega
   \]

2. **Combine the next series resistors:**
   - The result \( R_{\text{eq}} \approx 0.545 \, \Omega \) is now in series with the \( 1 \, \Omega \) resistor. Series resistors are added directly:
   \[
   R_{\text{total}} = 0.545 \, \Omega + 1 \, \Omega = 1.545 \, \Omega
   \]

3. **Now deal with the parallel combination:**
   - The result \( 1.545 \, \Omega \) is in parallel with the \( 4 \, \Omega \) resistor. Using the parallel formula again:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{1.545} + \frac{1}{4}
   \]
   \[
   \frac{1}{R_{\text{eq}}} \approx 0.647 + 0.25 = 0.897
   \]
   Therefore, the equivalent resistance of this section is:
   \[
   R_{\text{eq}} \approx \frac{1}{0.897} = 1.114 \, \Omega
   \]

4. **Series combination with the remaining resistors:**
   - Now, this result \( 1.114 \, \Omega \) is in series with the \( 7 \, \Omega \) and \( 5 \, \Omega \) resistors. Add them directly:
   \[
   R_{\text{total}} = 1.114 \, \Omega + 7 \, \Omega + 5 \, \Omega = 13.114 \, \Omega
   \]

5. **Final series combination with \( R_2 \) and \( R_1 \):**
   - Now, this result is in series with \( R_2 = 0.5 \, \Omega \) and \( R_1 = 7.2 \, \Omega \). Add them directly:
   \[
   R_{\text{final}} = 13.114 \, \Omega + 0.5 \, \Omega + 7.2 \, \Omega = 20.814 \, \Omega
   \]

### Final Answer:
The equivalent resistance of the circuit is approximately \( \boxed{20.8 \, \Omega} \).

Let me know if you want more details or have any questions!

---
Here are five follow-up questions you might want to explore:
1. How do you compute the equivalent resistance for more complex parallel networks?
2. What happens to the total resistance if one resistor in a parallel section is removed?
3. How would adding an additional resistor in series affect the total resistance?
4. Can you explain the difference between series and parallel circuits in terms of current and voltage?
5. What are some practical applications of combining resistors in series and parallel?

**Tip:** When simplifying resistive networks, always simplify the parallel sections first, then handle the series parts, step by step!

Find the equivalent resistance in Ohms (Ω) for the network below from the point of view of the terminals on the far left, where R1 = 7.2 Ω and R2 = 0.5 Ω.

Learn how to calculate the equivalent resistance of a complex circuit involving both series and parallel resistors, with R1=7.2Ω and R2=0.5Ω. Step-by-step solution provided for students at the high school or introductory college level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation