Photo AI
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 uses the graph by identifying the y-intercept, which represents the emf (electromotive force) of the solar cell. The graph shows that the emf is 0.70 V when the current is 0 mA. The slope of the line, which indicates the internal resistance, is found by determining the change in pd (potential difference) over the change in current. The gradient of the line can be calculated as the negative of the internal resistance, which in this case is 8.0 Ω.
Step 2
Deduce, using calculations, whether the circuits in Figure 5 and Figure 6 are suitable for this application.
Answer
For Figure 5, the total resistance in the circuit is 6.0 Ω, and with a required current of 75 mA, the voltage across the circuit can be calculated using Ohm's Law:
Since this voltage is less than the emf of 0.70 V, the circuit is suitable.
For Figure 6, the total resistance is also 6.0 Ω, but we need to consider additional internal resistance of the solar cell. Thus, if we assume a series circuit, the effective resistance is 6.0 Ω + 8.0 Ω = 14.0 Ω. The voltage required remains 0.45 V but with the higher resistance, the current produced is insufficient, therefore this circuit may not be suitable.
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
The power required to operate the road sign can be calculated as follows:
The total resistance is 6.0 Ω, and the current needed is 75 mA (0.075 A). Thus:
Since the solar cells operate at an efficiency of 4.0%, the input power required from sunlight is given by:
P_{input} = rac{P_{output}}{Efficiency} = rac{0.03375}{0.04} = 0.84375 ext{ W}
The sunlight intensity can be determined from the area of the solar cell supply, which is 32 cm², or 0.0032 m²:
Intensity = rac{P_{input}}{Area} = rac{0.84375 ext{ W}}{0.0032 m^2} = 263.5 ext{ W m}^{-2}