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Resistance effecting current through Voltmeter

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    When you hook up a voltmeter to a circuit, you are providing another possible pathway for current—that is, through the voltmeter. Usually this is not a problem, provided that the resistance of the voltmeter is much larger than the resistance of the circuit. Suppose you have a circuit containing a DC supply (voltage V) and a resistor R, and you want to use a voltmeter (with resistance 10 MΩ or 107 Ω) to measure the voltage drop across a resistance R. What is the largest value of R for which the current drawn from the DC supply will change by no more than 1%? (Hint: the current through R alone is I=V/R, where V is the DC supply voltage. What is the current through the parallel combination of R and the voltmeter?)


    2. Relevant equations
    V=IR
    Itotal=I1+I2
    I=V/R


    3. The attempt at a solution
    I really have no idea how to go about this. Only having one number really throws me for a loop. Right now I'm still trying to figure out what equations I need to use.
     
  2. jcsd
  3. Apr 16, 2009 #2

    S_Happens

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    Gold Member

    You have more numbers than you think, but you have to infer them from the problem. You have the resistance of the voltmeter (maybe, because "10 Mohms or 107 Ohms is not clear), and you know that you want the current through the resistor (with voltmeter in parallel) to be 99% I total (really think about the percentage as it's the key). What does that tell you about what the current through the voltmeter? From there you work can work backwards, because they are set up in parallel.

    If I give you anything else, then I will have basically answered the question.
     
  4. Apr 16, 2009 #3
    Oops, forgot to fix that, it's supposed to be 10 Mohms or 107Ohms. ^^; Sorry 'bout that.
    The current through the voltmeter will be 1 percent of the total current, right? So.. I need to find the current.
    so if I set up the equation I=V/R as I=IR/R...I'll end up with one. That's not right. I guess I'm not sure how the resistance of the voltmeter helps me at all with this.
    all the things I can think of to do have like, two unknowns.
    Because I know 1/Req=1/R1+1/R2 but I only know one of the resistances.
     
  5. Apr 16, 2009 #4

    S_Happens

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    Gold Member

    You don't need an actual number for the current to solve this problem. You know that Itotal is the current through the voltmeter plus the current through the resistor. You know that you want the current through the resistor to be a certain percentage of the total current. You tried an equation and set the currents equal (you didn't denote any difference between them with I1 and I2, so there's nothing for them to do except cancel each other out), but if one current is 99% of the total and the other is 1% then you know that can't be correct.

    Man, it's hard not to completely give it away, but I'll try again to give some hints.
    Start with the equation V=IR for each resistor in parallel
    1) What variable in the equation V=IR is always equivalent for both sides in parallel (not only in this example, but any)?
    2) Is there a way to plug the 99% and 1% into Itotal = I1 + I2?

    Using those two hints and the resistance of the voltmeter, you can turn two equations of V=IR (6 variables) into one equation that can solve for the other resistance. The last hint is that you need to figure out when a variable represented by a letter is REALLY a variable.
     
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