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    the question is attached. i got as far as saying that R2 is twice R1 (i.e. R2 = 2R1). could someone explain how to do this?
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    If the current through A is zero, you can just treat the problem as if A didn't exist. So you can calculate the ratio of the two resistors easily.

    \epsilon=I \left( R_{{1}}+R_{{2}} \right)=15V
    v=I R_{{2}}=8V

    Which leads to
    {\frac {R_{{1}}+R_{{2}}}{R_{{2}}}}={\fr  ac {15}{8}}
    R_{{1}}={\frac {7}{8}}\,R_{{2}}

    Maybe this part will help you get to the solution.
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    (Original post by madima)
    the question is attached. i got as far as saying that R2 is twice R1 (i.e. R2 = 2R1). could someone explain how to do this?
    Do A have any resistance itself? It is not an ammeter, right?
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    It is an ammeter!
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    (Original post by tillhm)
    If the current through A is zero, you can just treat the problem as if A didn't exist. So you can calculate the ratio of the two resistors easily.

    \epsilon=I \left( R_{{1}}+R_{{2}} \right)=15V
    v=I R_{{2}}=8V
    erm. the current isn't the same for both cases, so i don't think you can use the same symbol for both. plus, the current, i, given in the question isn't the one running through both resistors, is it?

    (Original post by tillhm)
    Which leads to
    {\frac {R_{{1}}+R_{{2}}}{R_{{2}}}}={\fr  ac {15}{8}}
    R_{{1}}={\frac {7}{8}}\,R_{{2}}
    ... therefore you can't eliminate the current (I) like that.
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    (Original post by thomaskurian89)
    It is an ammeter!
    An ammeter in parallel! hmm a novelty.
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    (Original post by thomaskurian89)
    It is an ammeter!
    If it is an ammeter, it has negligible resistance. So a large current will pass it as the circuit is shorted by the ammeter, and R2 will have no effect on the circuit.
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    it isn't an ammeter. hence why the question words it as a "device".
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    (Original post by madima)
    it isn't an ammeter. hence why the question words it as a "device".
    So it must have a resistance, which I need to know.
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    (Original post by ssadi)
    So it must have a resistance, which I need to know.
    lol. i posted is the whole question.

    it's gonna end up being a simultaneous equation, and i'm sure about R2 being twice R1.
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    You know the current through it and the PD across it so ...
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    (Original post by thomaskurian89)
    It is an ammeter!
    Also the "v" is not clear in the diagram It would be helpful if the precise points between which v is needed is clarified, I am myopic for a fact.
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    (Original post by madima)
    lol. i posted is the whole question.

    it's gonna end up being a simultaneous equation, and i'm sure about R2 being twice R1.
    never mind
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    (Original post by madima)
    the question is attached. i got as far as saying that R2 is twice R1 (i.e. R2 = 2R1). could someone explain how to do this?
    I found the other equation:
    r2=8/(7/r1-0.001)
    Please give me a rep if it is correct.
    Workings:
    Resistance of A =8/0.001=8000ohm
    15-8=7
    current=7/r1
    i=7/r1
    I(r2)=7/r1-0.001
    Voltage=8v
    r2=8/(7/r1-.001)
    Thus the two simultaneous equations are found, you may solve them now.
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    (Original post by madima)
    erm. the current isn't the same for both cases, so i don't think you can use the same symbol for both. plus, the current, i, given in the question isn't the one running through both resistors, is it?
    The two equations above are for the first case (no current through) A. The equations are wrong no matter what though. I misread the question.

    (Original post by ssadi)
    Also the "v" is not clear in the diagram It would be helpful if the precise points between which v is needed is clarified, I am myopic for a fact.
    The potential difference v is clearly marked in the diagram.
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    I hope this is understandable and correct
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