Answer the following: How to connect three resistors each of resistance 8 Ω, so that the equivalent resistance of the combination is 12 Ω ? Draw diagram of the combination and justify your answer.
Generated by claude-sonnet-4-6 · 2026-06-15 06:42 · grounding rag
Model Answer
Connection: Connect two resistors in parallel, then connect the third resistor in series with this parallel combination.
Diagram:
```
┌──[8Ω]──┐
A ──────┤ ├──────[8Ω]──── B
└──[8Ω]──┘
(parallel pair) (series)
```
Justification:
Equivalent resistance of the two resistors in parallel:
$$\frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} \implies R_p = 4 \ \Omega$$
Now this $R_p$ is connected in series with the third resistor:
$$R_{eq} = R_p + 8 = 4 + 8 = \mathbf{12 \ \Omega}$$
Hence the equivalent resistance of the combination is 12 Ω.
Source: Chapter 11, Section 11.6 – Resistance of a System of Resistors
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Explanation
- The key idea is to use a mixed (series-parallel) combination. Two resistors in parallel give half resistance (4 Ω); adding the third in series brings it back up to 12 Ω.
- Examiners expect: correct identification of the combination type, a clear circuit diagram, and step-by-step calculation using both formulas ($1/R_p$ for parallel and $R_s = R_1 + R_2$ for series).
- Always verify your final answer matches the required value.