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Ammeters and voltmeters in series

  1. Jun 11, 2017 #1
    1. The problem statement, all variables and given/known data

    An ammeter and a voltmeter are joined in series to a cell.Their readings are A and V respectively.If a resistance is now joined in parallel with the voltmeter,
    A)both A and V will increase
    B)both A and V will decrease
    C)A will decrease,V will increase
    D)A will increase,V will decrease

    2. Relevant equations

    Basic theory behind ammeters and voltmeters:
    ▪Ammeter:Used for measuring current.In essence,it is a galvanometer (current detecting device) attached to a very small resistance attached to it in parallel(small so that it doesn't hinder the current it is measuring and gives an accurate reading)

    ▪Voltmeter:Used for measuring potential difference.In essence,it is a galvanometer with a very high
    resistance attached to it in series (High so that it doesn't allow current to flow through it,which would change the amount of current across the element being measured, and correspondingly change the potential difference across it,preventing us from getting an accurate reading)

    3. The attempt at a solution

    An ideal voltmeter has a resistance of infinity,so when it is connected with a cell in series,along with an ammeter,the only path for current to traverse the entire circuit would be through the voltmeter,but the high resistance wouldn't allow it to flow.

    So,the initial readings according to me ought to read:
    [Assuming emf of cell to be E]
    A=0 (no current through the circuit)
    V=E (potential across its two ends)

    Now,as soon as we connect a resistor through it,it provides the current a path to flow,so now we have a current flowing through the circuit.
    So,final readings according to me should be:
    [Assuming resistance connected to be R]

    A=E/R(which is obviously greater than before)
    V=?HERE LIES MY PROBLEM?

    Isn't the potential difference across the two ends same as before??
    Answers D by the way,so hopefully my method of deducing A was right(If it wasn't,plz do inform me)
    But,I am stuck on V.
     
    Last edited by a moderator: Jun 12, 2017
  2. jcsd
  3. Jun 11, 2017 #2

    phinds

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    Yes, as long as you assume the source is an ideal one with no internal resistance. Why do you see this as a problem?

    Any time you get stuck on this kind of problem you should draw a circuit diagram and label it will all known conditions (and if you have, as you do here, two different set of conditions, draw it twice).
     
  4. Jun 11, 2017 #3
    Coz the answer given is D:A increases,whereas V decreases.There isn't any option for V remains same,which is bugging me.
     
  5. Jun 11, 2017 #4

    phinds

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    So, you need to decide. Am I wrong or is the question using an unstated assumption?
     
  6. Jun 12, 2017 #5

    cnh1995

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    What is V? And what is E?
     
  7. Jun 12, 2017 #6
    There's no E...A and V are the readings on the ammeter and voltmeter respectively.
     
  8. Jun 12, 2017 #7

    cnh1995

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    E is the emf of the cell.
    What will remain unchanged? V or E?
     
  9. Jun 12, 2017 #8
    According to me,V
     
  10. Jun 12, 2017 #9

    cnh1995

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    Why? And why not E?
     
  11. Jun 12, 2017 #10
    Okk,yeah,I get your point,the potential difference across the terminals of a battery change as soon as current flows through it.(If the battery isn't ideal,of course)
     
  12. Jun 12, 2017 #11

    cnh1995

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    So either the battery isn't ideal or the meters aren't ideal. (Or both aren't ideal).
    Then D sounds correct to me.
     
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