A solar cell generates a potential difference of 0.1185 V when a 500 resistor is

Question

A solar cell generates a potential difference of 0.1185 V when a 500 resistor is connected across it, and a potential difference of 0.1730 V when 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 light energy to thermal energy in the 1000 external resistor? (Give answer in %. Do not enter unit)

Explanation / Answer

Let emf be E and internal resistance be r
then,
i = E/(R+r)
V = i*R
V = E*R/(R+r)

0.1185 = E*500 / (500+r) ....eqn 1
0.1730 = E*1000 / (1000+r) ....eqn 2

divide eqn 2 by eqn 1:
0.1730/0.1185 = 1000*(500+r) / (500*(1000+r))
1.46 = (500000+1000r) / ((500000+500r))
730000 + 730r = 500000+1000r
r = 851.85 ohm
Answer: internal resistance = 851.85 ohm
--------------------------------------------------------
put value of r in eqn 1
0.1185 = E*500 / (500+r)   
0.1185 = E*500 / (500+851.85)
E = 0.320 V
= 320 mV
Answer: emf of battery is 320 mV
---------------------------------
received energy = 2mw/cm^2 * 5 cm^2 = 10 mw
converted energy = V*I
= V*E/(R+r)
=0.1730* (0.320 / (1000 + 851.85))
=2.99*10^-5 w
= 0.0299 mw

percentage= 0.0299*100 / 10
=0.3 %
Answer: 0.3 %

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