series circuit

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

  1. tonym1069

    Thread Starter New Member

    Aug 17, 2010
    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
    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 =