InterviewSolution
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InterviewSolution
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A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:A. (A) `(15//2)A`B. (B) `5 sqrt(3)A`C. (C) `5 sqrt(5)A`D. (D) `15 A` |
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Answer» Correct Answer - B Given `l=5+10 sin omega t ` `I_(eff)=[(int_(0)^(T)(i^2)dt)/(int_(0)^(T)dt)] = [ 1/T int_(0)^(T) (5 + 10 sin omega t)^(2) dt ]^(1//2)` `[ 1/T int_(0)^(T) (25 + 100 sin omega t)+100 sin^(2)omega t)]^(1//2)` But as `(1)/(T) int_(0)^(T) sin omega t dt = 0` and ` 1/T int_(0)^(T) sin^(2) omega t dt = 1/2` so `I_(eff) = [ 25 + 1/2 xx 100 ]^(1//2) = 5 sqrt(3)A`. |
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