1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measuring Capacitance given a voltmeter + ammeter

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the following:
    • a battery,
    • a digital voltmeter (which records voltage as a function of time),
    • a digital ammeter (which records current as a function of time),
    • a battered industrial capacitor,
    • two copper plates,
    • and several bits of copper wire.

    measure the capacitance of the industrial
    capacitor

    (there's a second segment of the question, but i understand it)

    2. Relevant equations
    C=Q/V. etc.
    I=dQ/dt.


    3. The attempt at a solution

    I have attempted to write a solution to this, but am only confusing myself.
    The voltage is a given, however I'm uncertain how I'm supposed to obtain Q from the dQ/dt graph, in order to determine capacitance.
    someone please shed some light -.-
     
  2. jcsd
  3. Sep 17, 2009 #2
    This is just an assumption that I thought: Hope it helps
    The ammeter is recording current as a function of time, so the reading is actually I/t.
    From here we can calculate Q.
    Q = I from ammeter readings . t
    the I is the reading from the ammeter, so it becomes Q = I/t . t
    the t cancel out so Q = I
    Does it makes sense?
     
  4. Sep 17, 2009 #3
    the usual way to measure the capacitance works with an ac voltage source!

    Your way is to vague!
     
  5. Sep 17, 2009 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Have you had integrals in calculus class?

    Well, not really. The ammeter reading is I, period.

    It's not "his way", it's the way given in the problem statement and he is stuck with it. :smile:
     
  6. Sep 17, 2009 #5
    You answered this yourself with the provided equations.

    Differentiate C = Q/V
    Bearing in mind that I = dQ/dt as well as the nature of a capacitance of a capacitor

    Then look at what you are provided and it's very simple. Note the function of time parts of the question.

    P.S. aren't you leaving this a bit late ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Measuring Capacitance given a voltmeter + ammeter
  1. Ammeter and Voltmeter (Replies: 2)

Loading...