How many minimum number of `2Omega` resistance can be connected to have an effective resistance of `1.5Omega`

Options

A. 3
B. 2
C. 4
D. 6

Explanation

To find the minimum number of `2Ω` resistors needed to achieve an effective resistance of `1.5Ω`, we can explore different combinations of resistors in series and parallel.

### Step-by-Step Solution:

1. **Understanding Series and Parallel Combinations**:
- When resistors are connected in series, their effective resistance (R_total) is the sum of their resistances:
\[ R_{\text{total}} = R_1 + R_2 + R_3 + \ldots \]
- When resistors are connected in parallel, the effective resistance is given by:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \]

2. **Checking with 1 Resistor**:
- If we use 1 resistor of `2Ω`, the effective resistance is `2Ω`. This is greater than `1.5Ω`.

3. **Checking with 2 Resistors**:
- If we connect 2 resistors in series:
\[ R_{\text{total}} = 2Ω + 2Ω = 4Ω \]
- If we connect 2 resistors in parallel:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{2Ω} + \frac{1}{2Ω} = \frac{2}{2Ω} \Rightarrow R_{\text{total}} = 1Ω \]
- Neither combination yields `1.5Ω`.

4. **Checking with 3 Resistors**:
- If we connect 3 resistors in series:
\[ R_{\text{total}} = 2Ω + 2Ω + 2Ω = 6Ω \]
- If we connect 3 resistors in parallel:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{2Ω} + \frac{1}{2Ω} + \frac{1}{2Ω} = \frac{3}{2Ω} \Rightarrow R_{\text{total}} = \frac{2}{3}Ω \approx 0.67Ω \]
- If we connect 2 in parallel and 1 in series:
- Parallel combination of 2 resistors:
\[ R_{\text{parallel}} = 1Ω \]
- Adding the third resistor in series:
\[ R_{\text{total}} = 1Ω + 2Ω = 3Ω \]
- None of these combinations yield `1.5Ω`.

5. **Checking with 4 Resistors**:
- If we connect 4 resistors in series:
\[ R_{\text{total}} = 2Ω + 2Ω + 2Ω + 2Ω = 8Ω \]
- If we connect all 4 in parallel:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{2Ω} + \frac{1}{2Ω} + \frac{1}{2Ω} + \frac{1}{2Ω} = \frac{4}{2Ω} \Rightarrow R_{\text{total}} = 0.5Ω \]
- Now, let's connect 3 resistors in series and 1 in parallel:
- Series combination of 3 resistors:
\[ R_{\text{series}} = 2Ω + 2Ω + 2Ω = 6Ω \]
- Now, connect this combination in parallel with the 4th resistor:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{6Ω} + \frac{1}{2Ω} \]
\[ \frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} \Rightarrow R_{\text{total}} = \frac{6}{4} = 1.5Ω \]

Thus, the minimum number of `2Ω` resistors required to achieve an effective resistance of `1.5Ω` is **4**.