The correct combination of three resistances `1Omega, 2Omega and 3Omega` to get equivalent resistance `(11)/(5)Omega` is

Options

A. All three are combines in paralle
B. All three are combines in paralle
C. `1Omega and 2Omega` in parallel and `3Omega` is in series to both
D. `2Omega and 3Omega` in parallel and `3Omega` is in series to both

Explanation

To find the correct combination of three resistances (1Ω, 2Ω, and 3Ω) that yields an equivalent resistance of \( \frac{11}{5} \)Ω, we will analyze the possible combinations of these resistors in series and parallel.

### Step-by-Step Solution:

1. **Understanding the Combinations**:
- We can combine resistors in series and parallel.
- The formula for resistors in series is:
\[ R_{\text{series}} = R_1 + R_2 + R_3 \]
- The formula for resistors in parallel is:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

2. **Testing All Combinations**:
- We will check different combinations of the given resistors to see which one gives us \( \frac{11}{5} \)Ω.

3. **Combination 1: All in Parallel**:
- For \( R_1 = 1Ω, R_2 = 2Ω, R_3 = 3Ω \):
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \]
\[ \frac{1}{R_{\text{eq}}} = 1 + 0.5 + 0.333 = \frac{11}{6} \]
\[ R_{\text{eq}} = \frac{6}{11}Ω \quad \text{(not the correct answer)} \]

4. **Combination 2: 1Ω and 2Ω in Parallel, 3Ω in Series**:
- For \( R_1 = 1Ω, R_2 = 2Ω \) in parallel and \( R_3 = 3Ω \) in series:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{1} + \frac{1}{2} = 1 + 0.5 = \frac{3}{2} \]
\[ R_{\text{parallel}} = \frac{2}{3}Ω \]
\[ R_{\text{eq}} = R_{\text{parallel}} + R_3 = \frac{2}{3} + 3 = \frac{2}{3} + \frac{9}{3} = \frac{11}{3}Ω \quad \text{(not the correct answer)} \]

5. **Combination 3: 2Ω and 3Ω in Parallel, 1Ω in Series**:
- For \( R_1 = 2Ω, R_2 = 3Ω \) in parallel and \( R_3 = 1Ω \) in series:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]
\[ R_{\text{parallel}} = \frac{6}{5}Ω \]
\[ R_{\text{eq}} = R_{\text{parallel}} + R_3 = \frac{6}{5} + 1 = \frac{6}{5} + \frac{5}{5} = \frac{11}{5}Ω \quad \text{(correct answer)} \]

### Conclusion:
The correct combination of resistances to achieve an equivalent resistance of \( \frac{11}{5} \)Ω is to connect the 2Ω and 3Ω resistors in parallel and then connect the 1Ω resistor in series with them.