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An engineer wants to use solar cells to provide energy for a filament lamp in a road sign - AQA - A-Level Physics - Question 4 - 2017 - Paper 1
Question 4
An engineer wants to use solar cells to provide energy for a filament lamp in a road sign. The engineer first investigates the emf and internal resistance of a sola... show full transcript
Worked Solution & Example Answer:An engineer wants to use solar cells to provide energy for a filament lamp in a road sign - AQA - A-Level Physics - Question 4 - 2017 - Paper 1
Step 1
Explain how the engineer uses the graph to obtain the values for the emf and internal resistance of the solar cell.
Answer
The engineer reads the graph to find the y-intercept, which represents the electromotive force (emf) of the solar cell, determined to be 0.70 V. The slope of the line is calculated by taking two points from the graph, which gives the value of internal resistance. The gradient of the graph represents the change in potential difference for a given change in current, and the internal resistance can be deduced using Ohm's law, where the gradient is the negative internal resistance.
Step 2
Deduce, using calculations, whether the circuits in Figure 5 and Figure 6 are suitable for this application.
Answer
For circuit in Figure 5:
- Total resistance = Road Sign Resistance + Internal Resistance;
- Total Resistance = 6.0 Ω + 8.0 Ω = 14.0 Ω.
- Current (I) = Voltage (V) / Total Resistance = 0.70 V / 14.0 Ω = 0.050 A (50 mA).
For circuit in Figure 6:
- Total resistance remains the same as before: 14.0 Ω.
- Current (I) = 0.70 V / 14.0 Ω = 0.050 A (50 mA).
Neither circuit provides the necessary 75 mA of current, therefore both circuits are unsuitable for this application.
Step 3
Calculate the minimum intensity, in W m−2, of the sunlight needed to provide the minimum current of 75 mA to the road sign when it has a resistance of 6.0 Ω.
Answer
First, calculate the total power needed for current of 75 mA:
- Using Ohm's law: P = I²R, where I = 0.075 A and R = 6.0 Ω,
- Power (P) = (0.075 A)² × 6.0 Ω = 0.03375 W, or 33.75 mW.
Next, considering efficiency:
- If efficiency is 4%, then Output Power = 0.04 × Input Power.
- Therefore: Input Power = Output Power / Efficiency = 0.03375 W / 0.04 = 0.84375 W.
To find the intensity (solar power per unit area):
- Total area of the solar cell = 32 cm² = 0.0032 m².
- Intensity = Power / Area = 0.84375 W / 0.0032 m² = 263.6 W m−2 (rounded to 264 W m−2).