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Direct current

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data
    when the balance point is achieved , there're no current passing thru , so Galvanometer should show 0 . Am i right... so there's no current passing thru the 20 ohm resistor . so my working is ( 8.1/120 )(40) = 2.7 V ...
    correct me if i'm wrong....


    2. Relevant equations


    3. The attempt at a solution
    V_xy = V_ac = (9/10) x 10 = 8.1v
    k = 8.1/120
    V_xy = k (40) = 2.7v

    (5/ (5+20+2) ) X E " = 2.7 V

    E" = 14.58v
     

    Attached Files:

  2. jcsd
  3. Nov 19, 2014 #2

    BvU

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    What is the problem statement ?
    No. It only means there is no current passing through the galvanometer.
     
  4. Nov 19, 2014 #3

    BvU

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    No current passing through the galvanometer means there is no potential difference over the thing. If we call the point between the 5 ##\Omega## and the 20 ##\Omega## resistors point z: VAC = VXZ.

    For VAC you have the right result (but perhaps a typo?) : VAC / VAB = 40/120 and VAB = 9V * 9/10.

    And for E' (single quote in the drawing) you also have the right result.
     
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