Answer :
Answer:
Solved
Explanation:
consider the emf of solar cell be ε and the output voltage be V . thus
V = ε - i r
= ε - ( V / R ) r
potential difference of 0.10 V when a 500 resistor is connected across the solar cell
0.1 V = ε - ( 0.1 V / 400Ω ) r .......(1)
potential difference of 0.15 V when a 1200 resistor is connected across the solar cell
0.15 V = ε - ( 0.15 V / 1200Ω ) r ..........(2)
on solving equations (1) and (2)
a) internal resistance r = 400 Ω
b) emf of the solar cell is ε = 0.2 V
c) received energy from light is P received = 2.0 mW/cm^2
= ( 2.0 mW/cm^2 ) ( 4 cm^2 )
= 8 mW
efficiency of the cell for converting the light energy to thermal energy in the 1200 external resistor
η = ( 0.15)^2 / 1200×8×10^-3 W
= 0.234%
The internal resistance, the emf of the solar cell, the efficiency are mathematically given as
R=2.538*10^2
e=0.184V
n=0.127%
The internal resistance, the emf of the solar cell and the efficiency of the cell
a)
Generally the equation for the voltage is mathematically given as
V=e-iR
Therefore
V1=e(V1/V2)R
0.10=e-(0.10/300)R....1
0.15=e(0.15/1100)R....2
Therefore
R=2.538*10^2
b)
e=0.10+(0.10/300)R
e=0.184V
c)
efficiency n
n=(v^2/R)
[tex]n=\frac{(0.15^2/1100)*10^4}{8*10^-4*2*10^{-3}}[/tex]
n=1.278*10^{-3}
n=0.127%
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