The resultant resistance of two resistance in series is 50 `Omega` and it is 12 ` Omega` , when they are in parallel. The individual resistances are

Options

A. `20 Omega and 15 Omega`
B. `15 Omega and 30 Omega`
C. `20 Omega and 30 Omega`
D. `10 Omega and 15 Omega`

Answer

To find the individual resistances \( R_1 \) and \( R_2 \) given that their resultant resistance in series is 50 ohms and in parallel is 12 ohms, we can follow these steps:

### Step 1: Set up the equations
For resistors in series, the total resistance \( R_s \) is given by:
\[ R_s = R_1 + R_2 \]
Given that \( R_s = 50 \, \Omega \), we have:
\[ R_1 + R_2 = 50 \quad \text{(1)} \]

For resistors in parallel, the total resistance \( R_p \) is given by:
\[ R_p = \frac{R_1 \times R_2}{R_1 + R_2} \]
Given that \( R_p = 12 \, \Omega \), we can substitute \( R_1 + R_2 \) from equation (1):
\[ 12 = \frac{R_1 \times R_2}{50} \quad \text{(2)} \]

### Step 2: Rearranging equation (2)
From equation (2), we can rearrange it to find \( R_1 \times R_2 \):
\[ R_1 \times R_2 = 12 \times 50 = 600 \quad \text{(3)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( R_1 + R_2 = 50 \)
2. \( R_1 \times R_2 = 600 \)

We can express \( R_1 \) in terms of \( R_2 \) from equation (1):
\[ R_1 = 50 - R_2 \quad \text{(4)} \]

Substituting equation (4) into equation (3):
\[ (50 - R_2) \times R_2 = 600 \]
Expanding this gives:
\[ 50R_2 - R_2^2 = 600 \]
Rearranging leads to:
\[ R_2^2 - 50R_2 + 600 = 0 \quad \text{(5)} \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( R_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -50, c = 600 \):
\[ R_2 = \frac{50 \pm \sqrt{(-50)^2 - 4 \times 1 \times 600}}{2 \times 1} \]
Calculating the discriminant:
\[ R_2 = \frac{50 \pm \sqrt{2500 - 2400}}{2} \]
\[ R_2 = \frac{50 \pm \sqrt{100}}{2} \]
\[ R_2 = \frac{50 \pm 10}{2} \]
This gives us two possible values for \( R_2 \):
\[ R_2 = \frac{60}{2} = 30 \quad \text{or} \quad R_2 = \frac{40}{2} = 20 \]

### Step 5: Find corresponding \( R_1 \)
Using equation (4) to find \( R_1 \):
1. If \( R_2 = 30 \):
\[ R_1 = 50 - 30 = 20 \]
2. If \( R_2 = 20 \):
\[ R_1 = 50 - 20 = 30 \]

Thus, the individual resistances are:
\[ R_1 = 30 \, \Omega, \quad R_2 = 20 \, \Omega \quad \text{or vice versa.} \]

### Final Answer:
The individual resistances are \( R_1 = 30 \, \Omega \) and \( R_2 = 20 \, \Omega \).

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