Three resistances `4Omega` each of are connected in the form of an equilateral triangle. The effective resistance between two corners is

Options

A. `8 Omega`
B. `12 Omega`
C. `(3)/(8) Omega`
D. `(8)/(3)Omega`

Answer

To find the effective resistance between two corners of an equilateral triangle formed by three resistances of 4 ohms each, we can follow these steps:

### Step 1: Understand the Configuration
We have three resistors, each of 4 ohms, connected in the shape of an equilateral triangle. Let's label the corners of the triangle as A, B, and C. The resistors are connected as follows:
- Resistor R1 between A and B
- Resistor R2 between B and C
- Resistor R3 between C and A

### Step 2: Identify the Resistors in the Circuit
To find the effective resistance between two corners, say A and B, we need to analyze the circuit:
- The resistor R1 (4 ohms) is directly between A and B.
- The resistors R2 (4 ohms) and R3 (4 ohms) are connected to point C.

### Step 3: Calculate the Equivalent Resistance
When looking from point A to point B:
- The resistor R1 is in series with the combination of R2 and R3, which are in parallel.

#### Step 3.1: Calculate the Equivalent Resistance of R2 and R3 in Parallel
The formula for the equivalent resistance \( R_{eq} \) of two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R2} + \frac{1}{R3} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
Thus, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = 2 \, \text{ohms} \]

#### Step 3.2: Add the Series Resistance
Now, we add the resistance R1 (4 ohms) to the equivalent resistance of R2 and R3 (2 ohms):
\[ R_{total} = R1 + R_{eq} = 4 + 2 = 6 \, \text{ohms} \]

### Step 4: Final Calculation
The effective resistance between points A and B is therefore:
\[ R_{AB} = 6 \, \text{ohms} \]

### Conclusion
The effective resistance between two corners of the equilateral triangle formed by three 4-ohm resistors is **6 ohms**.

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