How Do I Find the Internal Resistance in this Circuit?

[arkofnoah]

Homework Statement

http://img408.imageshack.us/img408/6537/screenshot20100808at134.png

Homework Equations

The Attempt at a Solution

I redrew the circuit as follow:

http://img293.imageshack.us/img293/6247/screenshot20100808at135.png

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.

 

[ehild]

The ideal millivoltmeter has infinitely high resistance.

ehild

 

[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 what does the 5.0mV mean? what does the change of 2.0mV mean? any clue?

 

[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

 

[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?

 

[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?

 

[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

 

[arkofnoah]

oh okay that's more of a typo

thanks for your help!