1.

The average value of alternating current `I=I_(0) sin omegat` in time interval `[0, pi/omega]` isA. `(2I_(0))/pi`B. `2I_(0)`C. `(4I_(0))/pi`D. `I_(0)/pi`

Answer» Correct Answer - A
`I_(av)=(underset(0)overset(pi//omega)intIdt)/(pi/omega-0)=omega/piunderset(0)overset(pi//omega)intI_(0) sin omegatdt=omega/pi[(I_(0)(-cos omegat))/omega]_(0)^(pi//omega)=-omega/piI_(0)/omega[cos pi-cos 0]=-I_(0)/pi[-1-1]=(2I_(0))/pi`


Discussion

No Comment Found

Related InterviewSolutions