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The e.m.f. developed by a photovoltaic cell can be taken as proportional to the logarithm of the intensity of radiation impinging on it. For 10 W/ m2 radiation, a cell develops an e.m.f. of 0.33 V and drives a current of 2.2 mA into a 100Ω load. Calculate: (2000)
(a) The open-circuit voltage at 25 W/m2.
(b) The internal resistance of the cell.
Correct answer is '50'. Can you explain this answer?
Verified Answer
The e.m.f. developed by a photovoltaic cell can be taken as proportio...
We know that e0 is directly proportional to logarithmic at I
i.e. e0 α log10 I
where I = 10 W/m2
e0 = K log10 10
=> 0.33 = K
(a) Given that I = 25 W/m2
e0 = K log10 I
e0 = K log10 25
= 0.33 log10 25
= 0.33 x 1.4 = 0.461 V
(b) R = 100Ω
I = 2.2 mA
V = IR
= 2.2 x 100 = 0.22 V
Internally dropped potential
= (0.33 - 0.22) V = 0.11V
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Most Upvoted Answer
The e.m.f. developed by a photovoltaic cell can be taken as proportio...
We know that e0 is directly proportional to logarithmic at I
i.e. e0 α log10 I
where I = 10 W/m2
e0 = K log10 10
=> 0.33 = K
(a) Given that I = 25 W/m2
e0 = K log10 I
e0 = K log10 25
= 0.33 log10 25
= 0.33 x 1.4 = 0.461 V
(b) R = 100Ω
I = 2.2 mA
V = IR
= 2.2 x 100 = 0.22 V
Internally dropped potential
= (0.33 - 0.22) V = 0.11V
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The e.m.f. developed by a photovoltaic cell can be taken as proportional to the logarithm of the intensity of radiation impinging on it. For 10 W/ m2 radiation, a cell develops an e.m.f. of 0.33 V and drives a current of 2.2 mA into a 100Ω load. Calculate: (2000)(a) The open-circuit voltage at 25 W/m2.(b) The internal resistance of the cell.Correct answer is '50'. Can you explain this answer?
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