The equivalent resistance of two resistor connected in series is `6 Omega` and their equivalent resistance is `(4)/(3)Omega`. What are the values of resistances ?

Options

A. `4 Omega, 6 Omega`
B. `8 Omega, 1 Omega`
C. `4 Omega, 2 Omega`
D. `6 Omega, 2 Omega`

Answer

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

### Step 1: Write down the equations for series and parallel resistances.
For resistors in series:
\[ R_s = R_1 + R_2 \]
Given that \( R_s = 6 \, \Omega \), we have:
\[ R_1 + R_2 = 6 \quad \text{(Equation 1)} \]

For resistors in parallel:
\[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
Given that \( R_p = \frac{4}{3} \, \Omega \), we can substitute \( R_1 + R_2 \) from Equation 1:
\[ \frac{R_1 R_2}{6} = \frac{4}{3} \]
Multiplying both sides by 6 gives:
\[ R_1 R_2 = 8 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations.
Now we have two equations:
1. \( R_1 + R_2 = 6 \)
2. \( R_1 R_2 = 8 \)

From Equation 1, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 6 - R_1 \]

### Step 3: Substitute \( R_2 \) into Equation 2.
Substituting \( R_2 \) into Equation 2:
\[ R_1(6 - R_1) = 8 \]
Expanding this gives:
\[ 6R_1 - R_1^2 = 8 \]
Rearranging the equation:
\[ R_1^2 - 6R_1 + 8 = 0 \]

### Step 4: Solve the quadratic equation.
We can solve this quadratic equation using the quadratic formula:
\[ R_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -6, c = 8 \):
\[ R_1 = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ R_1 = \frac{6 \pm \sqrt{36 - 32}}{2} \]
\[ R_1 = \frac{6 \pm \sqrt{4}}{2} \]
\[ R_1 = \frac{6 \pm 2}{2} \]
This gives us two possible values:
\[ R_1 = \frac{8}{2} = 4 \quad \text{or} \quad R_1 = \frac{4}{2} = 2 \]

### Step 5: Find \( R_2 \).
Using \( R_1 = 4 \):
\[ R_2 = 6 - R_1 = 6 - 4 = 2 \]
Using \( R_1 = 2 \):
\[ R_2 = 6 - R_1 = 6 - 2 = 4 \]

### Conclusion:
Thus, the values of the resistances are:
\[ R_1 = 4 \, \Omega \quad \text{and} \quad R_2 = 2 \, \Omega \]

### Final Answer:
The resistances are \( 4 \, \Omega \) and \( 2 \, \Omega \).

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