series circuit

Discussion in 'Homework Help' started by tonym1069, Feb 2, 2012.

  1. tonym1069

    Thread Starter New Member

    Aug 17, 2010
    1
    0
    Hello,
    I am self teaching myself. I picked up a problem online and need help.

    An arc lamp takes 9.6A at 55V. It is operated from a 120V supply. Find the value of the stabiling resistor to be connected in serries.

    So I tryed first 120V/9.6A=12.5 ohm. But then I thought a resistor must drop the voltage from 120V to 55V. So then I took voltage devider formula and solved for Rx using 12.5 ohms as R total. I got 5.729 ohms. The answer for the problem is 6.77 ohm.

    Thank you for your help.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,098
    Well, you are very close to a right solution. But you mix something in your calculations.

    R_tot = 120V/9.6A = 12.5Ω
    R_arc = 55V/9.6A = 5.72916667Ω


    So the resistor you need is equal to
    Rs = R_tot - Rarc = 6.770Ω

    But this problem can also be solved with the help of II Kirchoff's Law.

    Rs = ( 120V - 55V)/9.6A =
     
Loading...