Calculate EMF & Resistance of Solar Cell

[Amadeo]

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.

 

[lewando]

You were off to a good start by using the homework template in your previous post. Why stop now? You run the risk of having your thread become "disappeared" (depending on the patience level of the moderators).

That being said,

for is not the current in a circuit like this given by this equation i=ε/r+R?

More like i=ε/(r+R)

I tried using V=ε-(V/r+R)r

Did you mean: V=ε-(V/(r+R))r ? That implies that i = (V/(r+R)), which is not right-- i = V/R.

Check your math, post what you did.

EDIT:

so V1= ε(V1/R1)r
and V2=ε(V2/R2)r

I think you would rather mean to say:
V1 = ε - (V1/R1)r
V2 = ε - (V2/R2)r

 

[collinsmark]

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.

You really should define your terms regarding what term corresponds to what. But I'm assuming that
"V" is the measured voltage at the solar cell's terminals,
"ε" is the solar cell's emf,
"r" is the solar cell's internal resistance, and
"R" the external resistor.

If so, the first set of equations look correct to me. I.e., V = ε-ir; i = V/R → V = ε - (V/R)r.

But the second set of equations are each missing a minus sign. They are supposed to be based on the last equation just above. They should be:

V1 = ε - (V1/R1)r
V2 = ε - (V2/R2)r

Edit: Or using slightly better notation:
V_1 = \varepsilon - \frac{V_1}{R_1}r

V_2 = \varepsilon - \frac{V_2}{R_2}r

 

[collinsmark]

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.

And like @lewando said, something is not quite correct with your proposed V = \varepsilon - \frac{V}{r + R}r equation. It's almost correct, but there's a mistake using V vs \varepsilon (you're using one when you should be using the other). Again, be sure to carefully define your terms.

 

[Amadeo]

Thank you for your replies.

Yes, I made a mistake- I did mean to say

V1 = ε - (V1/R1)r
V2 = ε - (V2/R2)r

Actually, I originally tried to solve the problem by with i=ε/(r+R)

R1=500, R2=1000, V1=.1v, V2=.15V

V1 = ε - i1r

V2 = ε - i2r

i= ε/(r+R)

i1= ε/(r+R1)
i2 = ε/(r+R2)

so,

V1 = ε - (ε/(r+R1))r
V2 = ε - (ε/(r+R2))r

so

ε = (V1/(1-(r/(r+R))))

so

V2 = (V1/(1-(r/(r+R)))) - ((V1/(1-(r/(r+R))))/(r+R2))r

then I tried to solve for r (and got nonsense answers).

But, is it correct to substitute i1= ε/(r+R1) into V1 = ε - i1r? Because the solution manual is saying that i1=(V1/R) is to be substituted for i in V1 = ε - i1r. But, like I said, I am confused by that, because it seems that i1=ε/(r+R1).

The calculation is certainly easier if we take i1=V1/R as opposed to i1=ε/(r+R1).

 

[collinsmark]

Thank you for your replies.

Yes, I made a mistake- I did mean to say

V1 = ε - (V1/R1)r
V2 = ε - (V2/R2)r

Actually, I originally tried to solve the problem by with i=ε/(r+R)

R1=500, R2=1000, V1=.1v, V2=.15V

V1 = ε - i1r

V2 = ε - i2r

i= ε/(r+R)

i1= ε/(r+R1)
i2 = ε/(r+R2)

so,

V1 = ε - (ε/(r+R1))r
V2 = ε - (ε/(r+R2))r

so

ε = (V1/(1-(r/(r+R))))

so

V2 = (V1/(1-(r/(r+R)))) - ((V1/(1-(r/(r+R))))/(r+R2))r

then I tried to solve for r (and got nonsense answers).

But, is it correct to substitute i1= ε/(r+R1) into V1 = ε - i1r? Because the solution manual is saying that i1=(V1/R) is to be substituted for i in V1 = ε - i1r. But, like I said, I am confused by that, because it seems that i1=ε/(r+R1).

It is completely valid to use

V_1 = \varepsilon - \frac{\varepsilon}{r + R_1}r,

V_2 = \varepsilon - \frac{\varepsilon}{r + R_2}r

as your starting equations. So if you got nonsensical answers, you must have made a mistake in your math. But the approach is valid.

The calculation is certainly easier if we take i1=V1/R as opposed to i1=ε/(r+R1).

Yes, using the solution your solutions manual used does simplify the math a little. But both are valid ways to solve the problem.

 

[Amadeo]

Yes, using the solution your solutions manual used does simplify the math a little. But both are valid ways to solve the problem.

Ah yes, V2 = (V1/(1-(r/(r+R1)))) - ((V1/(1-(r/(r+R1))))/(r+R2))r works out to the same answer (1000) as using V=ε-(V/R)r.

I think it is also becoming clear to me as to why they are both valid ways of solving the problem: If we have the current given by ε/(r+R) it should be equal to V/R because the value V in the numerator is less than ε in the same degree to which R is less than R+r. The reduction in the numerator is compensated for by reduction in the denominator and via versa.

Using V/R is sort of like pretending that the internal resistance of the cell doesn't exist, and that the cell itself has an emf which equals the voltage it would have if it did have internal resistance.

Thanks again.