A filament lamp is connected to a battery in a simple circuit. The battery has an electromotive force (EMF) of 6 V. When the circuit is switched on, the lamp glows and the connecting wires become slightly warm. Explain why the lamp and the wires become warm, and describe how this affects the efficiency of the circuit.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A student notices that after a filament lamp has been switched on for several minutes, both the lamp and the connecting wires feel warm to the touch. The battery supplying the circuit has an electromotive force (EMF) of 6 V.
Model answer (5 marks)
The 6 V EMF of the battery pushes electrons through the circuit, transferring electrical energy to the charges.
The filament lamp and the connecting wires have resistance. As the electrons flow through this resistance, electrical energy is converted into thermal energy (Joule heating). The lamp glows because the heat is radiated as light, while the wires feel warm from the same heat loss.
The heat produced in the wires is not the intended useful output of the circuit; it is wasted energy.
Efficiency is defined as the ratio of useful output energy (light from the lamp) to the total input energy supplied by the battery.
Because a significant portion of the input energy is lost as heat in the lamp and the wires, the efficiency of the circuit is less than 1 (or less than 100 %).
The filament lamp and the connecting wires have resistance. As the electrons flow through this resistance, electrical energy is converted into thermal energy (Joule heating). The lamp glows because the heat is radiated as light, while the wires feel warm from the same heat loss.
The heat produced in the wires is not the intended useful output of the circuit; it is wasted energy.
Efficiency is defined as the ratio of useful output energy (light from the lamp) to the total input energy supplied by the battery.
Because a significant portion of the input energy is lost as heat in the lamp and the wires, the efficiency of the circuit is less than 1 (or less than 100 %).
Examiner tips
- Use the term ‘Joule heating’ to link resistance and heat. Show the energy flow: EMF → electrons → resistance → heat. Define efficiency as useful output/total input. Explain that heat loss reduces the ratio below 1.
Common mistakes
- Confusing EMF with voltage drop across the lamp. Failing to mention that heat in the wires is waste energy. Using vague terms like ‘energy loss’ without linking to resistance and efficiency.
Mark scheme (5 marks)
- The battery (EMF) transfers energy to the charges / electrons in the circuit
- The resistance of the lamp (and wires) causes energy to be transferred as heat / thermal energy
- The heat produced by the wires is waste energy / not the intended useful output of the circuit
- Efficiency is the ratio of useful output energy to total input energy
- Because energy is wasted as heat (in wires and as heat from the lamp), the efficiency of the circuit is less than 1 / less than 100%
Key terms in this question
electromotive force (EMF) · efficiency
Related
- All Edexcel A-Level Physics (9PH0) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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