# A solar cell that is 15% efficient in converting solar to electric energy produces an energy flow of 1.00 kW/\text(m)^2 when exposed to full sunlight. The cell has an area of 30.0 \text(cm)^2. (a) What is the power output of the cell, in watts?

## Two sections: What is the power output of the cell, in watts? If the power calculated in part (a) is produced at 0.45 V , how much current does the cell deliver?

Nov 23, 2016

Here's what I got.

#### Explanation:

Start by calculating the energy flow of the solar cell in kilowatts per square centimeter, ${\text{kW cm}}^{- 2}$. Use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 m"^2 = 10^4"cm}}^{2}}}}$

to find

$1.00 {\text{kW"/color(Red)(cancel(color(black)("m"^2))) * (1color(Red)(cancel(color(black)("m"^2))))/(10^4"cm"^2) = 1.00 * 10^(-4) "kW cm}}^{- 2}$

Now, you know that the total surface of the solar cell is equal to ${\text{30.0 cm}}^{2}$. Use this to calculate the total power processed by the cell from the incoming solar energy

30.0 color(Red)(cancel(color(black)("cm"^2))) * (1.00 * 10^(-4)"kW")/(1color(Red)(cancel(color(black)("cm"^2)))) = 3.00 * 10^(-3)"kW"

The cell is said to have an efficiency of 15%, which basically means that for every $\text{100 kW}$ of processed solar power, i.e. the power input, only $\text{15 kW}$ are converted to electric power, i.e. the power output.

In your case, this efficiency will produce a power output of

3.00 * 10^(-3)color(Red)(cancel(color(black)("kW input"))) * "15 kW output"/(100color(Red)(cancel(color(black)("kW input")))) = 4.5 * 10^(-4)"kW"

Finally, to convert this to watts, use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 kW" = 10^3"W}}}}$

You will end up with

$4.5 \cdot {10}^{- 4} \textcolor{R e d}{\cancel{\textcolor{b l a c k}{\text{kW"))) * "1 W"/(10^3color(Red)(cancel(color(black)("kW")))) = color(darkgreen)(ul(color(black)("0.45 W}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the efficiency of the solar cell.

For part (b), use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P = I \cdot V}}}$

Here

• $P$ is electric power
• $I$ is electric current
• $V$ is voltage

Rearrange to solve for $I$

$P = I \cdot V \implies I = \frac{P}{V}$

Keeping in mind that in terms of units you have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 W" = "1 A" * "1 V}}}}$

plug in your values to find

I = (0.45color(red)(cancel(color(black)("V"))) * "A")/(0.45color(red)(cancel(color(black)("V")))) = color(darkgreen)(ul(color(black)("1.0 A")))