The Total resistance when connected in series in `9 Omega` and when connected in parallel is `2 Omega` The value of two resistance are

Answer

To solve the problem, we need to find the values of two resistances \( R_1 \) and \( R_2 \) given that their total resistance in series is \( 9 \, \Omega \) and their total resistance in parallel is \( 2 \, \Omega \).

### Step-by-Step Solution:

1. **Understanding Series Resistance**:
When resistors are connected in series, the total resistance \( R_s \) is given by the sum of the individual resistances:
\[ R_s = R_1 + R_2 \]
According to the problem, this is equal to \( 9 \, \Omega \):
\[ R_1 + R_2 = 9 \quad \text{(Equation 1)} \]

2. **Understanding Parallel Resistance**:
When resistors are connected in parallel, the total resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]
According to the problem, this is equal to \( 2 \, \Omega \):
\[ \frac{1}{2} = \frac{1}{R_1} + \frac{1}{R_2} \quad \text{(Equation 2)} \]

3. **Rearranging Equation 2**:
We can rewrite Equation 2 in terms of \( R_1 \) and \( R_2 \):
\[ \frac{1}{2} = \frac{R_1 + R_2}{R_1 R_2} \]
Cross-multiplying gives:
\[ R_1 + R_2 = \frac{R_1 R_2}{2} \]
Substituting Equation 1 into this gives:
\[ 9 = \frac{R_1 R_2}{2} \]
Rearranging this leads to:
\[ R_1 R_2 = 18 \quad \text{(Equation 3)} \]

4. **Solving the System of Equations**:
Now we have two equations:
- \( R_1 + R_2 = 9 \) (Equation 1)
- \( R_1 R_2 = 18 \) (Equation 3)

We can express \( R_2 \) in terms of \( R_1 \) using Equation 1:
\[ R_2 = 9 - R_1 \]
Substituting this into Equation 3:
\[ R_1(9 - R_1) = 18 \]
Expanding this gives:
\[ 9R_1 - R_1^2 = 18 \]
Rearranging leads to:
\[ R_1^2 - 9R_1 + 18 = 0 \]

5. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ R_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -9, c = 18 \):
\[ R_1 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ R_1 = \frac{9 \pm \sqrt{81 - 72}}{2} \]
\[ R_1 = \frac{9 \pm \sqrt{9}}{2} \]
\[ R_1 = \frac{9 \pm 3}{2} \]
This gives us two possible values for \( R_1 \):
\[ R_1 = 6 \quad \text{or} \quad R_1 = 3 \]

6. **Finding Corresponding \( R_2 \)**:
If \( R_1 = 6 \):
\[ R_2 = 9 - 6 = 3 \]
If \( R_1 = 3 \):
\[ R_2 = 9 - 3 = 6 \]

### Final Answer:
The values of the two resistances are:
\[ R_1 = 6 \, \Omega, \quad R_2 = 3 \, \Omega \quad \text{(or vice versa)} \]