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Homework Help: 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


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    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.

  4. Oct 11, 2005 #3
    ok thank u sir!:smile:
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