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.

 

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 =

Textbook - https://www.allaboutcircuits.com/textbook/direct-current/chpt-5/simple-series-circuits/