Some students investigate electric circuits - Edexcel - GCSE Physics - Question 3 - 2012 - Paper 1
Question 3
Some students investigate electric circuits.
They set up this circuit.
What can the students do to increase the brightness of the lamp?
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Worked Solution & Example Answer:Some students investigate electric circuits - Edexcel - GCSE Physics - Question 3 - 2012 - Paper 1
Step 1
What can the students do to increase the brightness of the lamp?
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Answer
D decrease the resistance of the variable resistor.
Reducing the resistance of the variable resistor allows more current to flow through the circuit, which increases the brightness of the lamp.
Step 2
(b)(i) Complete the circuit below with an ammeter and a voltmeter correctly connected.
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Answer
To complete the circuit, ensure that an ammeter is placed in series with the lamp and a voltmeter is connected in parallel across the lamp. Both meters should be labeled correctly, ensuring clarity in function. The correct symbols for the ammeter and voltmeter should be used.
Step 3
(b)(ii) Calculate the potential difference (voltage) across the lamp.
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Answer
Using Ohm's law, the potential difference (V) can be calculated using the formula:
V=IimesR
Where:
I = 0.5 A (current)
R = 8 Ω (resistance)
Substituting the values:
V=0.5Aimes8Ω=4V
Thus, the potential difference across the lamp is 4 V.
Step 4
(c) Explain why only some of the electrical energy is transferred to light energy in the lamp.
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Answer
In a lamp, not all electrical energy is converted into light energy. Some energy is lost as heat due to resistance in the filament of the lamp. This heat energy dissipates into the surroundings and therefore does not contribute to light production. This is why lamps are often classified as incandescent, where a significant portion of energy is not converted to visible light.
Step 5
(d) Calculate the power being supplied to the lamp.
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Answer
Power (P) can be calculated using the formula:
P=IimesV
Where:
I = 0.4 A (current)
V = 5 V (potential difference)
Substituting the values:
P=0.4Aimes5V=2W
Thus, the power being supplied to the lamp is 2 W.
