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: Since \(75\ \Omega\) lies between one resistor's value (\(50\ \Omega\)) and two in series (\(100\ \Omega\)), try building it as one plain \(50\ \Omega\) resistor in series with some smaller combination contributing the remaining \(75-50 = 25\ \Omega\).

Step 2: Check whether two \(50\ \Omega\) resistors in parallel can supply exactly that remaining \(25\ \Omega\): \(\dfrac{1}{R} = \dfrac{1}{50}+\dfrac{1}{50} = \dfrac{1}{25}\), giving \(R = 25\ \Omega\), a match.

Step 3: Add this parallel pair's resistance in series with the third resistor: \(25 + 50 = 75\ \Omega\).
\[ R_{\text{total}} = \boxed{75\ \Omega} \]