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

Options

A. `4 Omega and 5 Omega`
B. `2 Omega and 7 Omega`
C. `3 Omega and 6 Omega`
D. `1 Omega and 8 Omega`

Answer

To solve the problem of finding the values of two resistors \( R_1 \) and \( R_2 \) given their total resistance in series is \( 9 \, \Omega \) and in parallel is \( 2 \, \Omega \), we can follow these steps:

### Step 1: Set up the equations
When resistors are connected in series, the total resistance \( R_s \) is given by:
\[ R_s = R_1 + R_2 \]
According to the problem, this gives us:
\[ R_1 + R_2 = 9 \, \Omega \quad \text{(1)} \]

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 gives us:
\[ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} \quad \text{(2)} \]

### Step 2: Manipulate the equations
From equation (1), we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 9 - R_1 \quad \text{(3)} \]

Now, substitute equation (3) into equation (2):
\[ \frac{1}{R_1} + \frac{1}{9 - R_1} = \frac{1}{2} \]

### Step 3: Clear the fractions
To eliminate the fractions, multiply through by \( 2R_1(9 - R_1) \):
\[ 2(9 - R_1) + 2R_1 = R_1(9 - R_1) \]
This simplifies to:
\[ 18 - 2R_1 + 2R_1 = 9R_1 - R_1^2 \]
So we have:
\[ 18 = 9R_1 - R_1^2 \]

### Step 4: Rearrange into standard form
Rearranging gives us:
\[ R_1^2 - 9R_1 + 18 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:
\[ R_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -9, c = 18 \):
\[ R_1 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ R_1 = \frac{9 \pm \sqrt{81 - 72}}{2} \]
\[ R_1 = \frac{9 \pm \sqrt{9}}{2} \]
\[ R_1 = \frac{9 \pm 3}{2} \]
This gives us two possible values for \( R_1 \):
\[ R_1 = \frac{12}{2} = 6 \, \Omega \quad \text{or} \quad R_1 = \frac{6}{2} = 3 \, \Omega \]

### Step 6: Find \( R_2 \)
Using equation (3):
- If \( R_1 = 6 \, \Omega \), then \( R_2 = 9 - 6 = 3 \, \Omega \).
- If \( R_1 = 3 \, \Omega \), then \( R_2 = 9 - 3 = 6 \, \Omega \).

Thus, the values of the resistors are:
\[ R_1 = 3 \, \Omega, \quad R_2 = 6 \, \Omega \quad \text{(or vice versa)} \]

### Final Answer
The values of the two resistors are \( 3 \, \Omega \) and \( 6 \, \Omega \).

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