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

For Figure 5, the circuit's total resistance is the sum of the road sign resistance (6.0 Ω) and the internal resistance of the solar cell (8.0 Ω), totaling 14.0 Ω. The current through the circuit can be calculated using Ohm's law:

$I = \frac{V}{R}$

Substituting in the values:

$I = \frac{0.70 \text{ V}}{14.0 \text{ Ω}} = 0.050 ext{ A} \ (50 ext{ mA})$

Since this is below the required minimum current of 75 mA, Figure 5 is unsuitable.

For Figure 6, the configuration allows for two solar cells in parallel. The effective emf remains 0.70 V, but the internal resistance halves:

$R_{total} = \frac{8.0 \text{ Ω}}{2} = 4.0 \text{ Ω}$

Thus, the total resistance in the circuit is 10.0 Ω and the current can be recalculated:

$I = \frac{0.70 \text{ V}}{10.0 \text{ Ω}} = 0.070 ext{ A} \ (70 ext{ mA})$

While higher than in Figure 5, this is still insufficient. Therefore, neither circuit is suitable for the application.

To find the minimum intensity, first, calculate the power needed to maintain the 75 mA current:

Using Ohm’s Law, $V = IR = 0.075 imes 6.0 = 0.45 ext{ V}$

Next, the required power is: $P = VI = 0.45 imes 0.075 = 0.03375 ext{ W}$

Considering the efficiency of the solar cell (4.0%), we find the input power: $P_{input} = \frac{P}{\text{efficiency}} = \frac{0.03375}{0.04} = 0.84375 ext{ W}$

The solar intensity can be calculated as: $I_{intensity} = \frac{P_{input}}{\text{Area}} = \frac{0.84375}{32 \text{ cm}^2} \times \frac{10000 \text{ cm}^2}{1 \text{ m}^2} = 263.4 ext{ W m}^{-2}$