1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

DC Circuit

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    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.
    untitled-3.png

    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


    2. Relevant equations
    V = IR


    3. 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
     
    Last edited: Jan 19, 2010
  2. jcsd
  3. Jan 18, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  4. Jan 19, 2010 #3
    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
     
  5. Jan 19, 2010 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  6. Jan 19, 2010 #5
    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
     
  7. Jan 19, 2010 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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.

    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.

    Yes, it looks so.

    ehild
     
    Last edited: Jun 29, 2010
  8. Jan 19, 2010 #7
    Hi ehild

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

    Thanks
     
  9. Jan 20, 2010 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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 modelled 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
     
    Last edited: Jun 29, 2010
  10. Jan 20, 2010 #9
    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
     
  11. Jan 20, 2010 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  12. Jan 20, 2010 #11
    I understand now. Thank you so much for your explanation ehild !! :smile:

    *p.s. : can you help me on my other question about phase difference of wave? i have new question there. thanks in advanced :P
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: DC Circuit
  1. DC circuits (Replies: 1)

  2. DC Circuits (Replies: 13)

  3. DC circuits (Replies: 3)

  4. R and I of DC circuit (Replies: 8)

Loading...