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!

Numerical problem about circuits

  1. Oct 10, 2005 #1
    PROBLEM:the current at the terminals of a certain current source is measured with an ammeter having an internal resistance [tex]R_i=10 ohms[/tex] and is found to be [tex]11.988 mA[/tex] ;adding a [tex]1.2 kilo -ohms[/tex] resistance between the source terminals causes the ammeter reading to drop to [tex]11.889 mA[/tex].Find [tex]i_s[/tex] and [tex]R_s[/tex]

    NOTE:[tex]i_s[/tex] and [tex]R_s[/tex] constitute a practical current source.i.e [tex]R_s[/tex] is the internalresistance of the source [tex]i_s[/tex]

    note:my answers don't quite match with the answer at the back of the textbook....
    my calulations: [tex]i_s=11.997 mA[/tex] and [tex]R_s=13200ohms[/tex]
    answers in the answer booklet:[tex]i_s=12mA[/tex] and [tex]R_s=10kilohms[/tex]:uhh:
     
  2. jcsd
  3. Oct 11, 2005 #2

    CarlB

    User Avatar
    Science Advisor
    Homework Helper

    I get the two voltages at the top of the ammeter as .11988V and then .11889V. The two currents are then 11.988mA and 11.889mA+.11889V/1.2kOhms.

    The change in voltage was .00099V
    The change in current was .000075mA = .000000075A

    So Rs = .00099V/.000000075A = 13,200 ohms just like you got. Maybe they're rounding to a single significant figure.

    Then Is = 11.988 mA + .11988V / 13.2 Kohms = 11.99708 mA

    Maybe your answers are right, except for the rounding.

    Carl
     
  4. Oct 11, 2005 #3
    ok thank u sir!:smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Numerical problem about circuits
  1. Circuit problem (Replies: 9)

  2. Circuits problem (Replies: 1)

Loading...