A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference of 0.31 V when a 1000 Ω resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is 4.0 cm2 and the rate per unit area at which it receives energy from light is 5.5 mW/cm2. What is the efficiency of the cell for converting light energy to thermal energy in the 1000 Ω external resistor?

a)

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

(1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when ,

Using Ohm's Law, the current is:

2) when ,

Using Ohm's Law, the current is:

Now we can rewrite eq.(1) in two forms:

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

is the voltage

is the current

is the internal resistance

Solving for E,

c)

In this part, we are told that the area of the cell is

While the intensity of incoming radiation (the energy received per unit area) is

This means that the power of the incoming radiation is:

This is the power in input to the resistor.

The power in output to the resistor can be found by using

where:

is the resistance of the resistor

is the current on the resistor (found in part A)

Susbtituting,

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr