Question

How are 50Ω resistors connected so as to give effective resistance of 75Ω.

• A. three resistors of 50Ω each in parallel
• B. three resistors of 50Ω each in series
• C. two resistors of 50Ω each in parallel
• D. two resistors of 50Ω each in parallel and the combination in series with another 50 Ω resistors.

Hint:

When resistors of equal value $R$ are in parallel, the resistance is $R/n$. For two $50\Omega$ resistors, it's $50/2 = 25\Omega$. Adding $50\Omega$ in series gives $25 + 50 = 75\Omega$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Resistors can be combined in series to increase total resistance ($R_s = R_1 + R_2$) or in parallel to decrease total resistance ($1/R_p = 1/R_1 + 1/R_2$). Mixed circuits allow for specific intermediate values.

Step 2: Detailed Explanation:
Let's evaluate option (D): 1. Parallel Part: Two $50\Omega$ resistors in parallel: \[ R_p = \frac{50 \times 50}{50 + 50} = \frac{2500}{100} = 25\Omega \] 2. Series Part: This $25\Omega$ combination is connected in series with another $50\Omega$ resistor: \[ R_{total} = R_p + 50 = 25 + 50 = 75\Omega \] This matches the required effective resistance.

Step 3: Final Answer:
To get $75\Omega$, two $50\Omega$ resistors should be in parallel, then connected in series with a third $50\Omega$ resistor.