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Circuits: What does the voltmeter read?

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A E = 12.0V battery with an internal resistance r = 1.80Ohm is connected to two 8.80kilo-Ohm resistors in series. An ammeter of internal resistance 0.410Ohm measures the current exiting the battery and at the same time a voltmeter with internal resistance 17.7kilo-Ohm measures the voltage across one of the 8.80kilo-Ohm resistors in the circuit. What does the ammeter read? What does the voltmeter read?

    2. Relevant equations

    V = IR
    Kirchhoff's Loop laws
    Resistors in parallel: ((1/R1)+(1/R2)+...)^-1 = R(equivalent)

    3. The attempt at a solution

    I drew the circuit described above and successfully found the ammeter reading of the current out of the battery ( or the total current). In my diagram (which should be right, given that I got I right), the battery, the ammeter, and the resistors are in series, while the voltmeter is in parallel with one of the resistors.

    My I(battery) was 0.000817 A, which I found using I = V/R after finding the total equivalent resistance.

    I'm having trouble finding the voltmeter reading, which should be measuring the voltage across one of the 8.8 kΩ resistors. I've done two separate calculations, but neither has been correct:

    1. V(through 8.8kΩ resistor) = I(battery)*8.8 ---> from V = IR
    2. Since the voltmeter has some internal resistance, some current could split along that path. I did Kirchhoff's loop equations to find the current through the resistor and used that current times the resistance to find V.


    Method 2 was wrong-- is this because I messed up my math, or because the theory was wrong and no current goes through the voltmeter?
     
  2. jcsd
  3. Mar 17, 2012 #2

    Delphi51

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    Homework Helper

    Method 2: Current goes through the 17.7 k and the 8.8 k in parallel so
    V = I*R = .000817*5878 = 4.80 V.
     
  4. Mar 17, 2012 #3

    PeterO

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    Homework Helper

    This voltmeter has a resistance only about twice the value of the resistor, so connecting it in parallel with one of them with reduce the effective resistance of that pair.

    You will effectively have a a battery in Series with an ammeter, in series with an 8.8 kΩ resistor in series with an effective (approx) 5.9 kΩ resistor.

    What is the PD across the 5.9 kΩ resistor then? That should be what the Voltmeter reads.

    [that 5.9 kΩ value I quote is approximate, you should be working with what you really get when a 8.8 kΩ and 17.7 kΩ resistor are connected in parallel.]
     
  5. Mar 18, 2012 #4
    Thanks to all!
     
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