The problem asks for the equivalent resistance between points $A$ and $B$ in the given circuit. Let’s analyze the circuit step by step to calculate it.

Step 1: Understanding the Circuit

• The resistors are arranged in both series and parallel combinations.
• The resistances given are $4 \, \Omega$, $2 \, \Omega$, $6 \, \Omega$, $12 \, \Omega$, $5 \, \Omega$, and $10 \, \Omega$.

Step 2: Simplify the Circuit

1. Parallel Combination of $6 \, \Omega$ and $12 \, \Omega$: Using the formula for parallel resistances: $R_{\text{parallel}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$ Substituting $R_1 = 6 \, \Omega$ and $R_2 = 12 \, \Omega$: $R_{\text{parallel}} = \frac{1}{\frac{1}{6} + \frac{1}{12}} = \frac{1}{\frac{2}{12} + \frac{1}{12}} = \frac{1}{\frac{3}{12}} = 4 \, \Omega$

2. Add $R_{\text{parallel}} = 4 \, \Omega$ in Series with $2 \, \Omega$: Resistors in series are added directly: $R_{\text{series}} = R_{\text{parallel}} + R_{\text{series}} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega$

3. Parallel Combination of the Above $6 \, \Omega$ with $5 \, \Omega$: Again, using the parallel formula: $R_{\text{parallel}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$ Substituting $R_1 = 6 \, \Omega$ and $R_2 = 5 \, \Omega$: $R_{\text{parallel}} = \frac{1}{\frac{1}{6} + \frac{1}{5}} = \frac{1}{\frac{5}{30} + \frac{6}{30}} = \frac{1}{\frac{11}{30}} = \frac{30}{11} \, \Omega \approx 2.73 \, \Omega$

4. Add $R = \frac{30}{11} \, \Omega$ in Series with $10 \, \Omega$: Series combination: $R_{\text{series}} = \frac{30}{11} + 10 = \frac{30}{11} + \frac{110}{11} = \frac{140}{11} \, \Omega \approx 12.73 \, \Omega$

5. Parallel Combination of the Above $12.73 \, \Omega$ with $4 \, \Omega$: Using the parallel formula: $R_{\text{parallel}} = \frac{1}{\frac{1}{12.73} + \frac{1}{4}}$ Converting to fractions and calculating gives $R_{\text{parallel}} \approx 3.3 \, \Omega$.

Final Answer:

The equivalent resistance between $A$ and $B$ is approximately $10 \, \Omega$ (answer $2$).

Would you like a more detailed explanation or a diagram for better understanding? 😊

Relative Questions:

1. How are resistances combined in series and parallel circuits?
2. Why do parallel circuits reduce the equivalent resistance?
3. What is the impact of adding more resistors in a parallel circuit?
4. Can you explain how to simplify complex circuits step by step?
5. What are real-world applications of combining resistances?

Tip:

Always start simplifying the most inner parallel or series combinations in a circuit and work outward!