The equivalent resistance of network of three `2Omega ` resistors can not be

Options

A. `0.67 `
B. `2 Omega `
C. `3 Omega `
D. `6 Omega `

Explanation

To determine the equivalent resistance of a network of three 2Ω resistors, we need to consider the different ways these resistors can be connected: in series and in parallel.

### Step 1: Calculate Equivalent Resistance in Series
When resistors are connected in series, the equivalent resistance (R_eq) is simply the sum of their resistances.

\[ R_{eq} = R_1 + R_2 + R_3 \]

Given that each resistor has a resistance of 2Ω:

\[ R_{eq} = 2Ω + 2Ω + 2Ω = 6Ω \]

### Step 2: Calculate Equivalent Resistance in Parallel
When resistors are connected in parallel, the formula for equivalent resistance is given by:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

Substituting the values of the resistors:

\[ \frac{1}{R_{eq}} = \frac{1}{2Ω} + \frac{1}{2Ω} + \frac{1}{2Ω} \]

Calculating this gives:

\[ \frac{1}{R_{eq}} = \frac{3}{2Ω} \implies R_{eq} = \frac{2}{3}Ω \approx 0.67Ω \]

### Step 3: Mixed Combination
Now, let's consider a mixed combination where two resistors are in parallel and one is in series with them.

1. **Calculate the equivalent resistance of two resistors in parallel:**

\[ \frac{1}{R_{parallel}} = \frac{1}{2Ω} + \frac{1}{2Ω} = \frac{2}{2Ω} = 1Ω \]

2. **Now add the third resistor in series:**

\[ R_{total} = R_{parallel} + R_3 = 1Ω + 2Ω = 3Ω \]

### Conclusion
From the calculations, we have found the following possible equivalent resistances:
- In series: 6Ω
- In parallel: 0.67Ω (or 2/3Ω)
- Mixed combination (2 in parallel, 1 in series): 3Ω

### Answer
The equivalent resistance of the network of three 2Ω resistors cannot be equal to any value other than those calculated. Therefore, the equivalent resistance cannot be:
- 1Ω
- 2Ω
- 4Ω
- 5Ω