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)=(int_(0)^(pi//omega)Idt)/((pi)/(omega-0))=(omega)/(2)int_(0)^(pi//omega)I_(0) sin omega tdt =(omega)/(pi)[(I_(0)(-cosomegat))/(omega)]_(0)^(pi//omega)=-(omega)/(pi)(I_(0))/(omega)[cos pi-cos 0]=-(I_(0))/(pi)[-1-1]=(2I_(0))/(pi)`


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