# 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?

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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?

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?

##### 1 Answer

Here's what I got.

#### Explanation:

Start by calculating the energy flow of the solar cell in *kilowatts per square centimeter*,

#color(blue)(ul(color(black)("1 m"^2 = 10^4"cm"^2)))#

to find

#1.00 "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 **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 **for every** *power input*, only *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

#color(blue)(ul(color(black)("1 kW" = 10^3"W")))#

You will end up with

#4.5 * 10^(-4) color(Red)(cancel(color(black)("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

#color(blue)(ul(color(black)(P = I * V)))#

Here

#P# is electric power#I# is electric current#V# is voltage

Rearrange to solve for

#P = I * V implies I = P/V#

Keeping in mind that in terms of units you have

#color(blue)(ul(color(black)("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")))#