then
P=V^2/R
=109^2/11.077ohm
=1072.59W
so drop is
1300 - 1072.59 = 227.41 W
It is not necessary to calculate the actual resistance since it was given that the resistance remains constant..
Stove burner output power drop = (power at 120V) - (power at 109V) = 1300W - {[(109V/120V)^2]*(1300W)} = 227.41 Watts
Looks right to me. Maybe you just forgot to subtract? THe question is asking about the drop, so you need:
1300 - 1073.26 = 224.74 W
Ah, Damon is right, I didn't check the arithmetic....
1300W-1073.26W=226.74W drop
what am I doing wrong?
During a "brownout", the power-line voltage drops from 120 V to 109 V. By how many watts does the thermal power output of a 1300 W stove burner drop, assuming the resistance remains constant?
(work)
R=V^2/P
=120^2/1300W
=11.077 ohm
then
P=V^2/R
=109^2/11.077ohm
=1073.26 W
