The average value of alternating current `I=I_(0) sin omegat` in time interval `[0, pi/omega]` is
A
`(2I_(0))/(pi)`
B
`2I_(0)`
C
`(4I_(0))/(pi)`
D
`(I_(0))/(pi)`
Text Solution
Verified by Experts
The correct Answer is:
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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