A solar cell generates a potential difference of 0.10 V when a 500- resistance is connected across it and a potential difference of 0.16 V when a 1000- resistance is substituted. We have to find (a) the internal resistance and (b) the emf of the solar cell. (c) The area of the cell is and the intensity of light striking it is . We have to find the efficiency of the cell for converting light energy to internal energy in the 1000- external resistor.

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(a)

Let the emf of the solar cell be V and its internal resistance be r . The solar cell generates a potential difference of 0.10 V when a 500- resistor is connected across it. From Ohm’s law, we have

The other data is that it generates a potential difference of 0.16 V when a 1000- resistance is connected across it. Using this data we write the second equation for determining the values of and r. It is

By dividing the two equations we get a linear equation, which can be solved for r. It is

(b)

The emf of the solar cell is

It is given that the area of the cell is and the intensity of light striking it is .

(c)

We will next find the efficiency of the cell for converting light energy to internal energy in the 1000- external resistor. Current flowing through the 1000- resistor connected to the solar cell is

Joule heat in the 1000- resistor will be

Light energy absorbed by the cell per second is

Therefore, the percentage efficiency of the solar cell for converting light energy into internal energy is