The voltage across a lamp is (6.0 +_ 0.1) V and the current flowing through it is (4.0 +_ 0.2) A. Find the power consumed with maximum permissible error in it?
ha ha scott great knowledge of error
http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm
http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf
Thanks Ravi.
100
as p is a
when quantities are multiplied , percent errors are added
(6.0 ± .17%) * (4.0 ± 5%) = ?
oops ... decimal error ... should be ... 6.0 ± 1.7%
sorry
6.1*4.2 = 25.62 max
5.9*3.8 = 22.42 min
difference between max and min = 3.2
divide by 2 = 1.6
nominal = 6*4 = 24
so
24 +/- 1.6
answer is
p=(24+_1.6)
As
p=vi
take log both side and
diffrentiate we get
(^=delta)
^p/p=^v/v+^i/i
^p=maximum possible error in p
^v=0.1
v=6
^i=0.2
i=4