Question

Choose the correct combination of three resistances \(1\ \Omega\), \(2\ \Omega\) and \(3\ \Omega\) to get equivalent resistance \(\dfrac{11}{5}\ \Omega\).

• A. All three are combined in parallel
• B. All three are combined in series
• C. \(1\ \Omega\) and \(2\ \Omega\) are in parallel and \(3\ \Omega\) is in series with both
• D. \(2\ \Omega\) and \(3\ \Omega\) are combined in parallel and \(1\ \Omega\) is in series with both

Hint:

Remember: \[ R_{\text{parallel}} = \frac{R_1R_2}{R_1+R_2} \] for two resistors in parallel.

• Series combination increases resistance
• Parallel combination decreases resistance

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Resistors can be combined in different networks to achieve specific total values. - In a series combination, the equivalent resistance is the sum of all parts: $R_s = R_1 + R_2 + \dots$ (the total is always larger than the largest single resistor). - In a parallel combination, the inverse values add up: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ (the total is always smaller than the smallest single resistor). - Mixed circuit combinations combine these two behaviors.
Step 2: Key Formula or Approach:
The target equivalent resistance is $R_{\text{eq}} = \frac{11}{5} \ \Omega = 2.2 \ \Omega$. Let's analyze the limits to eliminate incorrect choices quickly: - All series: $1 + 2 + 3 = 6 \ \Omega$ (Too large). - All parallel: smaller than $1 \ \Omega$ (Too small). This means we must test the mixed network options (C) and (D).
Step 3: Detailed Explanation:
Let's evaluate Option (D): 2 Ω and 3 Ω are connected in parallel, and this parallel pair is placed in series with the remaining 1 Ω resistor. 1. Calculate the parallel pair ($R_p$) for the 2 Ω and 3 Ω resistors: \[ R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} \ \Omega = 1.2 \ \Omega \] 2. Add the 1 Ω series resistor to this parallel group: \[ R_{\text{total}} = R_1 + R_p = 1 \ \Omega + \frac{6}{5} \ \Omega \] \[ R_{\text{total}} = \frac{5 + 6}{5} = \frac{11}{5} \ \Omega = 2.2 \ \Omega \] This configuration yields exactly the targeted value. Let's verify why Option (C) fails: - Parallel pair of 1 Ω and 2 Ω: $R_p = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \ \Omega$. - Adding 3 Ω in series: $3 + \frac{2}{3} = \frac{11}{3} \ \Omega \neq \frac{11}{5} \ \Omega$.
Step 4: Final Answer:
The correct combination is achieved when 2 Ω and 3 Ω are combined in parallel and 1 Ω is placed in series with both.