Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in (i) series, and (ii) parallel. Draw the arrangements of the resistors and find the equivalent resistance of each arrangement.
Generated by claude-sonnet-4-6 · 2026-06-15 06:43 · grounding rag
Model Answer
(i) Series combination:
$$\boxed{-\!\![\,2\,\Omega\,]\!-\!\![\,3\,\Omega\,]\!-\!\![\,6\,\Omega\,]\!-}$$
Using $R_s = R_1 + R_2 + R_3$:
$$R_s = 2 + 3 + 6 = \mathbf{11\ \Omega}$$
---
(ii) Parallel combination:
$$\begin{array}{c} -\!\![\,2\,\Omega\,]\!-\\ -\!\![\,3\,\Omega\,]\!-\\ -\!\![\,6\,\Omega\,]\!- \end{array}\ \text{(all connected between the same two points)}$$
Using $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$:
$$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1$$
$$\mathbf{R_p = 1\ \Omega}$$
Source: Chapter 11, Sections 11.6.1 and 11.6.2
---
Explanation
- Series formula: simply add all resistances; result is always greater than the largest individual value.
- Parallel formula: add the reciprocals; result is always less than the smallest individual value (here 1 Ω < 2 Ω ✓).
- Examiners expect the circuit diagram (even a simple line diagram), the correct formula written out, substitution shown, and the final answer with unit (Ω). Award split: ~1 mark for diagrams, 1 mark each for the two calculations.