A wire of resistance `18Omega` is divided into three equal parts. These parts are connected as sides of a triangle , the equivalent resistance of any two corners of the triangle will be

Options

A. `18Omega`
B. `9 Omega`
C. `6Omega`
D. `4Omega`

Answer

To solve the problem of finding the equivalent resistance between any two corners of a triangle formed by dividing a wire of resistance 18Ω into three equal parts, we can follow these steps:

### Step-by-Step Solution:

1. **Divide the Total Resistance**:
The total resistance of the wire is given as 18Ω. Since the wire is divided into three equal parts, the resistance of each part can be calculated as:
\[ R' = \frac{R}{3} = \frac{18Ω}{3} = 6Ω \]
Each of the three segments of wire has a resistance of 6Ω.

**Hint**: Remember that when a wire is divided into equal parts, the resistance of each part is the total resistance divided by the number of parts.

2. **Visualize the Triangle Configuration**:
We can visualize the three resistors (each 6Ω) as the sides of an equilateral triangle. We need to find the equivalent resistance between any two corners of the triangle.

3. **Identify the Configuration**:
Let’s label the corners of the triangle as A, B, and C. The resistors are:
- R1 (between A and B) = 6Ω
- R2 (between B and C) = 6Ω
- R3 (between C and A) = 6Ω

We want to find the equivalent resistance between points A and B.

4. **Combine Resistors**:
The resistors R2 and R3 (between B-C and C-A) are in parallel when looking from A to B:
\[ R_{parallel} = \frac{R2 \times R3}{R2 + R3} = \frac{6Ω \times 6Ω}{6Ω + 6Ω} = \frac{36Ω}{12Ω} = 3Ω \]

5. **Add the Series Resistance**:
Now, we have R1 (6Ω) in series with the equivalent resistance of R2 and R3 (3Ω):
\[ R_{AB} = R1 + R_{parallel} = 6Ω + 3Ω = 9Ω \]

6. **Final Calculation**:
The equivalent resistance between any two corners of the triangle (A and B) is therefore:
\[ R_{AB} = 9Ω \]

### Conclusion:
The equivalent resistance between any two corners of the triangle formed by the three equal parts of the wire is **9Ω**.