# Internal Resistance of a Real Battery

1. Oct 3, 2009

### Oijl

1. The problem statement, all variables and given/known data
A solar cell generates a potential difference of 0.10 V when a 400 ohm resistor is connected across it, and a potential difference of 0.15 V when a 900 ohm resistor is substituted.

What is the internal resistance of the solar cell?

2. Relevant equations
E - ir - iR = 0

3. The attempt at a solution
This solar cell is a real battery, in that it has an internal resistance. But I can imagine it as being an ideal battery with emf E connected in series to a resistor of resistance r, where r is the internal resistance of the real battery. So, with this cell connected (in series) to a 400 ohm resistor of resistance R, I can say, by the loop rule,
E - ir -iR = 0

The potential differences that the question says the real cell generates would be the potential across the ideal battery and the resistor r, such that
0.10 V = E - i_{1}r
where i_{1} is the current in the case where the 400 ohm resistor is in the circuit, and
0.15 V = E - i_{2}r
where i_{2} is the current in the case where the 900 ohm resistor is in the circuit.

And again, since E - ir - iR = 0, I can also write
0.1 - 400i = 0
and
0.15 - 900i = 0

So then I have four equations and four unknowns, and I can solve.

But when I do solve for r (which, remember, represents the internal resistance of the real cell), I get 120 ohms, and the program tells me this isn't correct.

I know I was verbose, so thanks for reading.

2. Oct 4, 2009

### Delphi51

Shouldn't that be E - 0.1 - 400i = 0 ?

You can easily get numerical values for the currents. In case one, applying I = V/R to the known resistor we have I = 0.1/400 = .00025 A
If you used those you would only have 2 equations to worry about.
I think your 120 ohm answer is a few times too small.

3. Oct 4, 2009

### Oijl

I thought at first that it should be E - 0.1 - 400i = 0.
Then I remembered that 0.1 V is the potential difference put out by the real battery. E represents the emf of the ideal battery, which combined with the imaginary resistor of the same resistance as the internal resistance of the real battery, would be the potential difference generated by the real battery. So we would have E - ir = 0.1.
And putting that into E - ir - iR = 0, we have 0.1 - iR = 0.

Hold on - does "connect across" mean in parallel? Because I've been treating the cell and the resistor as if they were in series. If they're in parallel, you're right, I could very simply say I = 0.1/400.

4. Oct 4, 2009

### Oijl

Yes, it worked. Thank you!

5. Oct 4, 2009

### Delphi51

Oh, you are quite right: If you use the .1 V as the potential of the real battery (E plus the iR) then you have only two elements to add: 0.1 - i*400 = 0. In fact that is the equation I used to find i.
But you still need to use all three elements to find E.

E, R and the 400 ohms are all in series.
So the sum around the loop has three potentials to add.
E - iR - i*400 = 0

Good idea to put in that i*400 = 0.1 from the first equation.
But also sub in for the other i so you have an equation with only the two variables E and R. Repeat all that for the 900 ohm case and get another one with the two variables.

I wonder if your mistake was keeping i as a variable, and not realizing that there are two different currents i1 and i2 for the two cases? Best to eliminate the i in each case.