Two resistors P (2 Ω) and Q (8 Ω) are connected in series to a battery. A student claims that Q will produce more heat per second than P. Is the student correct? Justify your answer with a calculation. How would the ratio of heat produced per second in P and Q change if the two resistors were instead connected in parallel to the same battery? Support your answer with a calculation.
Generated by claude-sonnet-4-6 · 2026-06-26 01:12 · grounding rag
Model Answer
Series Combination:
Total resistance = 2 + 8 = 10 Ω. Let battery voltage = V.
Current through both resistors: $I = V/10$
Heat produced per second (Power) = $I^2R$
- In P: $H_P = I^2 \times 2 = \frac{V^2}{100} \times 2 = \frac{2V^2}{100}$
- In Q: $H_Q = I^2 \times 8 = \frac{V^2}{100} \times 8 = \frac{8V^2}{100}$
Since Q has a higher resistance and current is the same in series, Q produces more heat per second. The student is correct.
Ratio $H_P : H_Q = 2 : 8 = \mathbf{1 : 4}$
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Parallel Combination:
Both resistors have the same potential difference V across them.
Heat produced per second = $V^2/R$
- In P: $H_P = V^2/2$
- In Q: $H_Q = V^2/8$
Ratio $H_P : H_Q = \frac{V^2}{2} : \frac{V^2}{8} = 4 : 1$
In parallel, P produces more heat per second than Q. The ratio reverses to 4 : 1.
Source: Chapter 11 – Electricity, Section 11.8 Electric Power
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Explanation
- In series, current is the same → use $P = I^2R$; the larger resistor dissipates more power.
- In parallel, voltage is the same → use $P = V^2/R$; the smaller resistor dissipates more power.
- Examiners want the formula stated, a numerical ratio calculated, and a clear conclusion for each case. Both parts are mandatory for full marks.