Two resistors are connected (a) in series (b) in parallel The equivalent resistance in the two cases are `9 ohm` and `2ohm` respectively. Then the resistances of the component resistor are

Options

A. `2 ohm` and `7 ohm`
B. `3 ohm` and `6 ohm`
C. `3 ohm` and `9 ohm`
D. `5 ohm` and `4 ohm`

Explanation

To solve the problem of finding the resistances of two resistors connected in series and parallel, we can follow these steps:

### Step 1: Set Up the Equations
Let the two resistors be \( R_1 \) and \( R_2 \).

1. When connected in series, the equivalent resistance \( R_s \) is given by:
\[ R_s = R_1 + R_2 \]
According to the problem, \( R_s = 9 \, \Omega \). Therefore, we have:
\[ R_1 + R_2 = 9 \quad \text{(Equation 1)} \]

2. When connected in parallel, the equivalent resistance \( R_p \) is given by:
\[ R_p = \frac{R_1 \cdot R_2}{R_1 + R_2} \]
According to the problem, \( R_p = 2 \, \Omega \). Therefore, we have:
\[ \frac{R_1 \cdot R_2}{R_1 + R_2} = 2 \quad \text{(Equation 2)} \]

### Step 2: Substitute Equation 1 into Equation 2
From Equation 1, we can express \( R_1 \cdot R_2 \):
\[ R_1 \cdot R_2 = R_p \cdot (R_1 + R_2) \]
Substituting \( R_p = 2 \) and \( R_1 + R_2 = 9 \):
\[ R_1 \cdot R_2 = 2 \cdot 9 = 18 \quad \text{(Equation 3)} \]

### Step 3: Solve the System of Equations
Now we have two equations:
1. \( R_1 + R_2 = 9 \) (Equation 1)
2. \( R_1 \cdot R_2 = 18 \) (Equation 3)

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

### Step 4: Solve the Quadratic Equation
We can solve the quadratic equation \( R_2^2 - 9R_2 + 18 = 0 \) using the quadratic formula:
\[ R_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -9, c = 18 \):
\[ R_2 = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
Calculating the discriminant:
\[ R_2 = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} \]
Thus, we have:
\[ R_2 = \frac{12}{2} = 6 \quad \text{or} \quad R_2 = \frac{6}{2} = 3 \]

### Step 5: Find \( R_1 \)
Using \( R_1 + R_2 = 9 \):
1. If \( R_2 = 6 \), then \( R_1 = 9 - 6 = 3 \).
2. If \( R_2 = 3 \), then \( R_1 = 9 - 3 = 6 \).

### Conclusion
The resistances of the two resistors are \( R_1 = 3 \, \Omega \) and \( R_2 = 6 \, \Omega \) (or vice versa).