renz.bagaporo
New member
- Joined
- Sep 1, 2012
- Messages
- 1
P = (E^2/ R). Approxiamte the maximum percent error in calculating for P if E = 120 and R = 2000, and the possible percent errors in measuring E and R are 3% and 4% respectively.
What is wrong with my solution:
delta E/ 120 = 0.03
delta E = 3.6
delta R/ 2000 = 0.04
delta R = 80
dP = (2E/R)dE + (-E^2/R^2) dR
= (2/2000)(120)(3.6) - (120^2/2000^2)(80)
= .144
P = (E^2) / R; P = (120^2 / 2000) = 7.2,
delta P / P = .144/ 7.2 = 0.02 x 100 = 2%?
the book says 10%. where have I gone wrong?
What is wrong with my solution:
delta E/ 120 = 0.03
delta E = 3.6
delta R/ 2000 = 0.04
delta R = 80
dP = (2E/R)dE + (-E^2/R^2) dR
= (2/2000)(120)(3.6) - (120^2/2000^2)(80)
= .144
P = (E^2) / R; P = (120^2 / 2000) = 7.2,
delta P / P = .144/ 7.2 = 0.02 x 100 = 2%?
the book says 10%. where have I gone wrong?