- #1
Amadeo
- 28
- 9
A solar cell generates a potential difference of .1v when a 500ohm resistor is connected across it, and a potential difference of .15v when a 1000ohm resistor is substituted. What is A.) the internal resistance of the solar cell and B.) the emf of the solar cell
I am a bit confused because, after some failed attempts, I resorted to the solution manual which said that the solution is found in the following way:
V= ε-ir; i=V/R → V=ε-(V/R)r
so V1= ε(V1/R1)r
and V2=ε(V2/R2)r
so the system can be solved.
But, it seems that this is a mistake, for is not the current in a circuit like this given by this equation i=ε/r+R? Are we not neglecting to account for the internal resistance of the cell by leaving out the r value in the denominator.
I tried using V=ε-(V/r+R)r and got some nonsense answers- but I don't know if this is because of mistakes in the laborious algebra, or because I set up the problem incorrectly.
I am a bit confused because, after some failed attempts, I resorted to the solution manual which said that the solution is found in the following way:
V= ε-ir; i=V/R → V=ε-(V/R)r
so V1= ε(V1/R1)r
and V2=ε(V2/R2)r
so the system can be solved.
But, it seems that this is a mistake, for is not the current in a circuit like this given by this equation i=ε/r+R? Are we not neglecting to account for the internal resistance of the cell by leaving out the r value in the denominator.
I tried using V=ε-(V/r+R)r and got some nonsense answers- but I don't know if this is because of mistakes in the laborious algebra, or because I set up the problem incorrectly.