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.

Answer 2

Each proposed arrangement of \(50\ \Omega\) resistors can be checked directly by computing its effective resistance and comparing it to the target of \(75\ \Omega\).

1. Three resistors of 50Ω each in parallel: For three equal resistors in parallel, \(\dfrac{1}{R} = \dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50} = \dfrac{3}{50}\), giving \(R = \dfrac{50}{3} \approx 16.7\ \Omega\), far below \(75\ \Omega\).
2. Three resistors of 50Ω each in series: For resistors in series, resistances simply add: \(50+50+50 = 150\ \Omega\), which overshoots the target substantially.
3. Two resistors of 50Ω each in parallel: \(\dfrac{1}{R} = \dfrac{1}{50}+\dfrac{1}{50} = \dfrac{2}{50}\), giving \(R = 25\ \Omega\), still well short of \(75\ \Omega\) with nothing left to add.
4. Two resistors of 50Ω each in parallel, combined in series with a third 50Ω resistor: The parallel pair gives \(25\ \Omega\) as computed above, and adding this in series with the third \(50\ \Omega\) resistor gives \(25+50 = 75\ \Omega\), matching the target exactly.

Only the mixed configuration, a parallel pair followed by a series resistor, produces the required effective resistance.

Therefore, the correct answer is two resistors of 50Ω each in parallel, and the combination in series with another 50Ω resistor.