please answer .what may be the answer. plus one physics
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Hey Mate
your answer is --
Given, % error in voltage V = 5%
& % error in current I = 2%
(i) we know that
R = V/I
=> ∆R/R ×100 = ( ∆V/V×100+∆I/I×100)
= ( 5%+2%) = 7%
therefore, % error in resistance is 7%
(ii) we know that
power P = VI
=> ∆P/P×100 = ( ∆V/V×100+∆I/I×100)
= ( 5%+2%) = 7%
Hence, % error in both resistance and power is 7%
HOPE IT HELP YOU
your answer is --
Given, % error in voltage V = 5%
& % error in current I = 2%
(i) we know that
R = V/I
=> ∆R/R ×100 = ( ∆V/V×100+∆I/I×100)
= ( 5%+2%) = 7%
therefore, % error in resistance is 7%
(ii) we know that
power P = VI
=> ∆P/P×100 = ( ∆V/V×100+∆I/I×100)
= ( 5%+2%) = 7%
Hence, % error in both resistance and power is 7%
HOPE IT HELP YOU
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