# Finding internal resistance

1. Aug 8, 2010

### arkofnoah

1. The problem statement, all variables and given/known data
http://img408.imageshack.us/img408/6537/screenshot20100808at134.png [Broken]

2. Relevant equations

3. The attempt at a solution
I redrew the circuit as follow:

http://img293.imageshack.us/img293/6247/screenshot20100808at135.png [Broken]

Is this correct (or even useful in the first place)?

Anyway when switch S1 is closed, the total resistance = rx + ry, the internal resistances of the batteries.

But when both switches are closed, I figured that the voltage increased by 0.02V, to 0.07V. However I am not sure this is the voltage across which components because the millimeter is connected in a rather strange way.

I'm lost from here onwards. I'm mostly troubled by the placement of the millivoltmeter I guess.

Am I on the right track? Any alternative method?

The answer is (A) by the way.

Last edited by a moderator: May 4, 2017
2. Aug 8, 2010

### ehild

The ideal millivoltmeter has infinitely high resistance.

ehild

3. Aug 8, 2010

### arkofnoah

okay i was thinking about that. when switch 1 is closed there is no current, and when both switches are closed the entire circuit system is basically a series circuit because no current flows though the millivoltmeter.

but i don't know what is the implication of this :tongue: what does the 5.0mV mean? what does the change of 2.0mV mean? any clue?

4. Aug 8, 2010

### ehild

When S2 is open the branch wit R just does not exist. Draw this arrangement. It might help. The voltage of what does the voltmeter read?

ehild

5. Aug 8, 2010

### arkofnoah

alright. so 5.0mV would be the potential difference between the two cells. Since there is no current flowing this is equal to the emf difference between the two cells, correct? so i can effectively treat X and Y as if it is a single cell with emf 5.0mV, right?

when s2 is closed, will the voltage increase by 2.0mV or decrease by 2.0mV? i know there is now current flowing in the branch with S2 but can you elaborate how exactly does this affect the millivoltmeter reading?

6. Aug 8, 2010

### arkofnoah

Oh okay i think i get it. So initially the voltmeter reading of 5.0mV is

$$E_{x} - E_{y}$$ where Ex and Ey is the emf of X and Y.

Now after closing the circuit with R, current flows through the internal resistor of X and the voltmeter reading of 3.0mV is

$$V_{x} - E_{y}$$ where Vx is the terminal p.d. of X, which is lower than Ex its emf.

The difference of these two is basically

$$E_{x} - V_{x} = Ir$$ where r is the internal resistance of battery

So 2mV = 5.0(r) and r is 0.004 ohm.

Am I correct?

7. Aug 8, 2010

### ehild

Perfect, except two minor errors: The first voltage reading is 50 mV and the other one differs from this by 20 mV.

From 2 mv =r (5 A) --->r = 0.0004 ohm, but you get the correct result if you use the original 20 mV.

ehild

8. Aug 8, 2010

### arkofnoah

oh okay that's more of a typo :tongue: