A solar cell generates a potential difference of 0.1111 V when a 500 resistor is

Question

A solar cell generates a potential difference of 0.1111 V when a 500 resistor is connected across it, and a potential difference of 0.1687 Vwhen a 1000 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 5.0 cm2, and the rate per unit area at which it receives energy from light is 2.0m W/ cm2. What is the efficiency of the cell for converting the light energy to thermal energy in the 1000 external resistor? (percent = Do not enter a unit)

Explanation / Answer

potential difference dV = E - I*r

curretn I = E/(R+r)

dV = E - E*r/(R+r) = E*(1 - r/(R+r)) = E*R/(R+r)

for R1 = 500 ohms   dV1 = 0.1111 v

0.1111 = E *500/(500+r) .............(1)

for R2 = 1000 ohms   dV2 = 0.1687 v

dV2 = E*1000/(1000+r)

0.1687 = E*1000/(1000+r)..........(2)

solving 1 & 2

0.1687/0.1111 = 2*(500+r)/(1000+r)

internal resistance r = 1076.6 ohm

===============

dV1 = E*R/(R+r)

E = dV1*(R+r)/R

E = 0.1111*(500+1076.6)/500

E = 0.35 v <<<<<------ANSWER

========================

rate of energy received = 2*5 = 10 W

rate of energy dissipated = V^2/R = 0.1687^2/1000 = 2.84*10^-5 W

efficiency = (2.84*10^-5/10)*100 = 0.0003 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.