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Meter circuit

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Meter in the circuit shown has a linear scale that has not been calibrated and the scale reading is 20 . When another resistor of resistance 2000 ohm is connected across XY . What is the scale reading of the meter

    2. Relevant equations



    3. The attempt at a solution

    Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005

    the reading given by the ammeter is 40600 times of the actual current .

    now with the resistor , I = 1.5/(2995 + 50+2000) , I = 0.0003

    the reading by the ammater will be 0.0003 x 40600 = 12

    but the answer given is 25
     

    Attached Files:

  2. jcsd
  3. Mar 27, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper


    Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005

    This is wrong. In the given circuit the total resistance is (2995+50 +2000)
    Find the current I.
    I = k*θ. where k is the constant of the meter and θ is the deflection in the mete.
    When you connect another 2000 ohm resistance in parallel with 2000 ohm resistance in the circuit, what is net resistance in the circuit? Find the current and equate it to k*θ'. find θ'.
     
  4. Mar 27, 2010 #3
    got it ! Thanks !
     
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