A solar cell that is \(15 \%\) efficient in converting solar to electric energy produces an energy flow of \(1.00 \mathrm{kW} / \mathrm{m}^{2}\) when exposed to full sunlight. (a) If the cell has an area of \(40.0 \mathrm{cm}^{2}\), what is the power output of the cell, in watts? (b) If the power calculated in part (a) is produced at \(0.45 \mathrm{V},\) how much current does the cell deliver?

The power input to the cell per unit area is given. To find the total power input to the cell, the power input per unit area needs to be multiplied by the total area of the cell. The cell's area is given in cm², but we need it in m² for the calculations. Convert cm² to m² using the conversion \(1 m^{2} = 10^{4} cm^{2}\). Hence, \(40.0 cm^{2} = 4.0 \times 10^{-3} m^{2}\). Next, calculate the total power input by multiplying it with the power per unit area, which is \(1.00 kW/m^{2} = 1.00 \times 10^{3} W/m^{2}\). Thus, the total power input \(P_{in} = 1.00 \times 10^{3} W/m^{2} \times 4.0 \times 10^{-3} m^{2} = 4.0 W\). Lastly, the power output \(P_{out}\) can be found by multiplying the total power input \(P_{in}\) with the efficiency (15%), which is \(P_{out} = 0.15 \times P_{in} = 0.15 \times 4.0 W = 0.60 W\).

For the second part of the problem, we're asked to find the current delivered by the cell. Using Ohm's law which states that the current \(I\) is equal to the power \(P\) divided by the voltage \(V\), \(I = P/V\). The power \(P\) is given from step 1 as 0.60 W, while the voltage \(V\) is given as 0.45 V. Hence, the current \(I = 0.60 W / 0.45 V = 1.33 A\).