Explain why the brightness of a lamp decreases when a second identical lamp is connected in series with it, compared to when the first lamp operates alone across the same power supply of negligible internal resistance.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
Connecting a second identical lamp in series increases the total resistance of the circuit.
Because the supply voltage is fixed and its internal resistance is negligible, the increased resistance causes the current through the circuit to decrease.
The voltage is now shared between the two lamps, so the potential difference across the original lamp is less than the full supply voltage.
The power dissipated in the original lamp is reduced (P = IV or P = V²/R), so it is less bright.
Because the supply voltage is fixed and its internal resistance is negligible, the increased resistance causes the current through the circuit to decrease.
The voltage is now shared between the two lamps, so the potential difference across the original lamp is less than the full supply voltage.
The power dissipated in the original lamp is reduced (P = IV or P = V²/R), so it is less bright.
Examiner tips
- Use the word ‘because’ to link cause and effect. Show the step that total resistance increases → current decreases → voltage drop per lamp decreases → power decreases. Mention the supply has negligible internal resistance to justify constant V.
- common_mistakes
- :
- Failing to state that the supply voltage is unchanged. Using the wrong formula for power (e.g. P = I²R without explaining the change in I). Not recognising that the voltage is divided between the two lamps.
Mark scheme (4 marks)
- Connecting a second lamp in series increases the total resistance of the circuit.
- Because the supply voltage is fixed (negligible internal resistance), the increased resistance causes the current through the circuit to decrease.
- The voltage is now shared/divided between the two lamps, so the potential difference across the original lamp is less than the full supply voltage.
- The power dissipated in the original lamp is reduced (P = IV or P = V²/R), so it is less bright.
Key terms in this question
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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