Question

Two resistances combine in series order to provide 50 ohms resultant resistance, and when they combine in parallel order, they provide 8 ohms resultant resistance. Then the value of each resistance is:

• A. 21 ohm and 29 ohm
• B. 10 ohm and 40 ohm
• C. 20 ohm and 30 ohm
• D. 15 ohm and 35 ohm

Hint:

To solve for resistances in series and parallel, use the formulas for the respective configurations and solve the system of equations.

Answer 1

The Correct Option is D

Let the resistances be \( R_1 \) and \( R_2 \). 1. The formula for two resistances in series: \[ R_{\text{series}} = R_1 + R_2 \] We are given: \[ R_{\text{series}} = 50 \, \text{ohms} \quad \Rightarrow \quad R_1 + R_2 = 50 \, \text{(Equation 1)} \] 2. The formula for two resistances in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \] We are given: \[ R_{\text{parallel}} = 8 \, \text{ohms} \quad \Rightarrow \quad \frac{1}{8} = \frac{1}{R_1} + \frac{1}{R_2} \] Multiplying both sides by \( R_1 R_2 \): \[ R_1 R_2 = 8(R_1 + R_2) \quad \Rightarrow \quad R_1 R_2 = 8 \times 50 = 400 \, \text{(Equation 2)} \] Now, solving these two equations: - Equation 1: \( R_1 + R_2 = 50 \) - Equation 2: \( R_1 R_2 = 400 \) We can solve this system by using the quadratic equation: \[ x^2 - (R_1 + R_2)x + R_1 R_2 = 0 \] \[ x^2 - 50x + 400 = 0 \] Solving this quadratic equation: \[ x = \frac{50 \pm \sqrt{50^2 - 4 \times 1 \times 400}}{2} = \frac{50 \pm \sqrt{2500 - 1600}}{2} = \frac{50 \pm \sqrt{900}}{2} = \frac{50 \pm 30}{2} \] Thus, \( x = 40 \) or \( x = 10 \). Therefore, the resistances are 10 ohms and 40 ohms. Thus, the correct answer is \( 10 \, \text{ohms} \) and \( 40 \, \text{ohms} \). % Correct Answer Correct Answer:} (B) 10 ohm and 40 ohm