Two resistors of 6 Ω and 3 Ω are first connected in series and then in parallel across the same 12 V battery. Find the ratio of total power dissipated in the series arrangement to that in the parallel arrangement.
Generated by claude-sonnet-4-6 · 2026-06-26 01:09 · grounding rag
Model Answer
Series combination:
$R_s = 6 + 3 = 9\ \Omega$
$P_s = \dfrac{V^2}{R_s} = \dfrac{12^2}{9} = 16\ \text{W}$
Parallel combination:
$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$, so $R_p = 2\ \Omega$
$P_p = \dfrac{V^2}{R_p} = \dfrac{144}{2} = 72\ \text{W}$
Ratio:
$$\frac{P_s}{P_p} = \frac{16}{72} = \boxed{\dfrac{2}{9}}$$
The ratio of power in series to parallel is 2 : 9.
Source: Chapter 11 – Electricity, Section 11.8 Electric Power
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Explanation
- Use $P = V^2/R$ since voltage (12 V) is the same in both cases — this is the quickest approach.
- Series resistance is simply the sum; for parallel, use the reciprocal formula.
- Examiners expect you to show both $R_s$ and $R_p$ calculations before writing the ratio. Do not skip steps in a 2-mark question.
- The key insight: parallel resistance is smaller, so more power is dissipated in parallel — hence the ratio is less than 1.