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Two resistors and voltmeter

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Two resistors are connected in series to an ideal battery. When the voltmeter is connected in parallel to one of the resistors, it showed 6V; when connected in parallel to another resistor it showed 4V. When the voltmeter was connected in parallel to both resistors (connected to pins A and B, see attachment) it showed 12V. What were the voltages across both resistors before the voltmeter was connected?

    2. Relevant equations

    Ohm's law

    3. The attempt at a solution

    I guess we can say from the third measurement that the voltage of the supply is 12V. Now the voltage across the first resistor is normally [itex] V_1 = V * \frac{R_1}{R_1 + R_2}[/itex], where V is the supply voltage. Although when the voltmeter is connected due to its internal resistance the voltage across 1st resistor changes; then the voltage across the voltmeter and the first resistor is [itex] V_1' = V * \frac{R_{01}}{R_{01}+R2}[/itex], and [itex] R_{01} = \frac{R_1 * r}{R_1 + r} [/itex], r is the voltmeter resistance. Thus [itex] V_1' = V * \frac{R_1r}{R_1r + R_2R_1 + R_2r}[/itex]. Same calculations can be made for the second resistor. Is that right? I can't figure out what to do next - any help appreciated.
     

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  3. May 3, 2014 #2

    BvU

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    So you have expressions for ##V_1'## and for ## V_2'##. You know the ratio is 4/6. What does that say about ##R_1/R_2## ?
     
  4. May 4, 2014 #3

    ehild

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    The question is what would be the voltages without the voltmeter. You wrote the expression for V1, you can write it in terms of R2/R1. On the same way, you can find V2 in terms of R1/R2.

    R1/R2 can be found from the ratio of the measured voltages, as BvU suggested.

    ehild
     
  5. May 4, 2014 #4
    Yes, thank you very much! Was quite close to the answer myself, I guess :)
     
  6. May 4, 2014 #5

    ehild

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    Yes, you were quite close, needed just a little push:smile:

    ehild
     
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