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

Concept: Equivalent resistance in:

• Series: \[ R=R_1+R_2+\cdots \]
• Parallel: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]

Step 1: Check option (D). First combine: \[ 2\ \Omega \text{ and } 3\ \Omega \] in parallel. Equivalent resistance: \[ R_p = \frac{2\times3}{2+3} \] \[ R_p = \frac{6}{5}\ \Omega \]

Step 2: Add the \(1\ \Omega\) resistance in series. \[ R_{\text{eq}} = 1+\frac{6}{5} \] \[ R_{\text{eq}} = \frac{5+6}{5} \] \[ R_{\text{eq}} = \frac{11}{5}\ \Omega \] This matches the required value. Therefore, the correct answer is: \[ \boxed{ 2\ \Omega \text{ and } 3\ \Omega \text{ in parallel, with } 1\ \Omega \text{ in series} } \]