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Reading of a Voltmeter

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data

    The circuit in the attached figure (excuse the terrible drawing, I suck at using the paint feature) all meters are idealized and the batteries have no appreciable internal resistance

    a) Find the reading of the voltmeter with the switch S open

    b) With the switch closed, find the reading of the voltmeter and the ammeter. Which way does the current flow through the switch?

    2. Relevant equations

    V = I/R (ohm's law)

    [tex]\Sigma[/tex]I = 0 (junction law)

    [tex]\Sigma[/tex]V = 0 (loop law)


    3. The attempt at a solution

    To find the reading of the voltmeter I attempted to find the currents going through the 2 emf's and then using the junction rule to find current at point a (the top) and then using ohm's law with that current and the 75 ohm resistor to find the reading of the voltmeter, with the switch open I treated that segment as if the switch line wasn't there, here's what I got

    for loop 1 (with the 25V emf)

    25V-I1(100 ohms) = 0

    25V = I1(100ohms), I1 = .25A

    for loop 2 (with the 15V emf)

    15V-I2(75 ohms) = 0

    15V = I2(75 ohms), I2 = .20A

    junction rule .25A-.20A-I = 0, I = .05A

    ohm's law for voltmeter

    I = V/R, V = IR -> V = (.05A)(75ohms) = 3.75V

    I'm not sure if I did the question correctly, if any one can check it for me I would appreciate it, thank you.
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    It looks like your attachment is coming through finally, so we can look at the circuit.

    With the switch open, there is only one loop containing two batteries in series and two resistors in series. That means there's only one current to solve for in part (a).

    The ammeter is on the open branch, so naturally it will read nothing. The voltmeter is measuring the potential "drop" across the 15-V battery and the 75-ohm resistor. So you'll need to find the voltage drop across the resistor, then "add in" the 15 volts from the battery appropriately.

    After that, we'll deal with part (b).
     
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