Determining the Internal Resistance of Battery X

[songoku]

Homework Statement

Two almost identical lead-acid accumulator batteries, X and Y, are connected with opposing polarities in a circuit which includes an ammeter, an ideal millivoltmeter and a resistor R as shown in figure below.

When S₁ is closed with S₂ left open, the millivoltmeter reads 50 mV. When S₂ is closed with S₁ remaining closed, the reading of the millivoltmeter changes by 20 mV and ammeter reads 5.0 A. What is the internal resistance of battery X ?

a. 4.0 x 10^-3
b. 6.0 x 10^-3
c. 10 x 10^-3
d. 14 x 10^-3
e. 16 x 10^-3

Homework Equations

V = IR

The Attempt at a Solution

Should we consider internal resistance of Y, because it is said "two almost identical lead-acid accumulator batteries" ? ( I think because it is almost identical, Y also has internal resistance that should be considered ).

After S₂ is closed, the reading changes by 20 mV --> I interpret this : the millivoltmeter reading becomes 30 mV.

Then I don't know...Thanks

 

[ehild]

Both batteries have internal resistance, and you have to take them equal,as it was not specified otherwise, and you can determine their sum only.

As they are lead-acid accumulator batteries of presumable of the same kind, the difference between then is their emf caused by the different amount of charge stored on each of them.

ehild

 

[songoku]

Hi ehild

I'm sorry there was mistake on the picture. I've fixed the circuit now. And I am still not able to do it...

Thanks

 

[ehild]

Well, it is OK then. The difference of the emf is Ey-Ex=0.05 V.
When S2 is closed, I =5 A flows through X, and no current through Y, as the voltmeter is ideal. It measures the sum of voltages across the batteries. The terminal voltage of X is less by I*rb than its emf. So the total voltage at the minus pole of battery X is more positive as before closing S2. The total voltage the voltmeter reads is -Ex+I*rb +Ey, the change in voltage is I*rb=0.02 V, I = 5 A, rb= ?

ehild

 

[songoku]

Why Ey-Ex=0.05 V? Should we consider the internal resistance of X and Y?

And I also don't understand this part :
"So the total voltage at the minus pole of battery X is more positive as before closing S2"

So basically, what we need is just the information 20 mV and the current 5 A?

Thanks

 

[ehild]

Why Ey-Ex=0.05 V? Should we consider the internal resistance of X and Y?

When S2 is open, no current flows as the voltmeter has infinite resistance. If there is no current, there is no drop of voltage across th einternal resistances. The voltmeter measures the potential difference between the points O and B. Assuming the potential zero at point O, it is -Ex at point A and increases by Ey from A to B, so it is Ey-Ex at B.

And I also don't understand this part :
"So the total voltage at the minus pole of battery X is more positive as before closing S2"

When S2 is closed, current flows in the bottom circuit. The current causes a potential drop across the internal resistance, so the voltage across the terminals of the battery x is lower than the emf. That is, the potential at A is less negative than it was before closing the switch S2.

So basically, what we need is just the information 20 mV and the current 5 A?

Yes, it looks so.

ehild

 

[songoku]

Hi ehild

If there is potential difference between O and B, does it mean that there is current flowing?

Thanks

 

[ehild]

Potential difference does not mean that current flows between the two points. But the bottom loop is closed, it contains a battery and finite resistances, so current flows out from the positive pole of the battery, through the ammeter, arriving at point A. There is a node at A, but no current can flow into the upper loop, as the infinite resistance of the voltmeter does not let it.

The real battery can be modeled as an ideal one with emf Ex and its internal resistance Ri in series. Now the negative pole of the real battery is connected to A.

So the whole current flows into the negative pole of the battery. It flows through Ri, causing a potential drop inside the battery and arrives to the point which is at potential -Ex with respect to O.

ehild

 

[songoku]

Hi ehild

I mean when S₂ is opened and S₁ closed. There is potential difference between O and B so will the current flow through the voltmeter from O to B? If not, why? I think current will flow if there is potential difference.

Thanks

 

[ehild]

Current flows between two points if these points are connected with a conductor.
If you have an isolated point charge, there is potential difference between two points in the electric field around the charge, if these points are at different distance from the charge. And no current.

A voltmeter takes on some current, but it is practically negligible. There are methods to measure voltage with no current at all.
When you have to solve a problem including voltmeters and ammeters, it is supposed that no current flows through the voltmeter and no voltage drops across the ammeter, unless the problem includes "real" meters, with some internal resistance.

In this problem, when S2 is opened and S1 is closed, there is no current flowing through the voltmeter. In reality, there is some, but you do not know it. It is very small.
Even the common cheap multimeters have internal resistance of about 50 Mega-ohm. If you would measure 50 mV with such a voltmeter, the current would be 10^-9A. Such current would cause a potential change of 4*10^-6 mV across the 4-ohm internal resistance of the battery. It is practically zero compared to the 50 mV measured.

ehild

 

[songoku]

I understand now. Thank you so much for your explanation ehild !

*p.s. : can you help me on my other question about phase difference of wave? i have new question there. thanks in advanced :P