Question

A solar cell generates a potential difference of 0.10 V when a 171.0 Ω resistor is connected across it, and a potential difference of 0.15 V when a 353.0 Ω resistor is substituted.
What is the internal resistance of the solar cell?

What is the emf of the solar cell?

The area of the cell is 1.80 cm², and the rate per unit area at which it receives energy from light is 7.00 mW/cm².
What is the efficiency of the cell for converting light energy to thermal energy in the 353.00 Ω external resistor? (Give answer in %. Do not enter unit)

Answer 1

Given

Potential difference (v1) = 0.1 v when the resistance (R1) = 171 ohms

Current (i1) = V1/R1 = 0.1 / 171 = 0.000584 A

Potential difference (v2) = 0.15 v when the resistance (R2) = 353 ohms

Current (i2) = V2/R2 = 0.15 / 353 = 0.000425 A

Let the emf of the cell be E and internal resistance be r

E – ir = V

E – 0.000584 r = 0.1 --------------(1)

E – 0.000425 r = 0.15 -------------(2)

Subtracting (1) from (2) we get

-0.000425r + 0.000584r = 0.15 – 0.1

0.000159 r = 0.05

r = 0.05 / 0.000159 = 314.46 ohms

E = 0.1 + 0.000584 * 314.46 = 0.283 v

Power dissipated (p1) = v^2 /R = 0.15^2 / 353 = 6.37 * 10^-5 W

Power received (p2) = 7 * 10^-3 * 1.80 = 0.0126 w

Efficiency = p2/p1 * 100 = 6.37 * 10^-5 * 100 / 0.0126 = 0.505 %