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. `(15//2)` ampB. `5sqrt(3)` ampC. `5sqrt(5)` ampD. `15`amp |
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Answer» Correct Answer - B Given `1=5+10 sin omegat` `I_(eff)=[(int_(0)^(T)I^(2)dt)/(int_(0)^(T) dt)]^(1//2)` `=[1/Tint_(0)^(T)(5+10 sin omega t)^(2) dt]^(1//2)` `=[1/Tint_(0)^(T) (25+100 sin omegat+100 sin^(2) omegat)]^(1//2)` But as `1/Tint_(0)^(T) sin omega t dt=0` and `1/Tint_(0)^(T) sin^(2) omega t dt=1/2` so `I_(eff)=[25+1/2xx100]^(1//2)` `=5sqrt(3)` amp |
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