Asked by Mimi
                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?
            
            
        Answers
                    Answered by
            ravi
            
    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
    
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
                    Answered by
            scott
            
    when quantities are multiplied , percent errors are added
(6.0 ± .17%) * (4.0 ± 5%) = ?
    
(6.0 ± .17%) * (4.0 ± 5%) = ?
                    Answered by
            ravi
            
    ha ha scott great knowledge of error
    
                    Answered by
            bobpursley
            
    http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm
    
                    Answered by
            bobpursley
            
    http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf
    
                    Answered by
            bobpursley
            
    Thanks Ravi.
    
                    Answered by
            scott
            
    oops ... decimal error ... should be ... 6.0 ± 1.7%
sorry
    
sorry
                    Answered by
            Damon
            
    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
    
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
                    Answered by
            Anonymous
            
    100
    
                    Answered by
            jackie
            
    as p is a
    
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