The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le (pi)/(omega)` `=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)` The average value of the voltage one cycle isA. `(V_(0))/(sqrt(2))`B. `(V_(0))/(2)`C. zeroD. `(2V_(0))/(pi)`

The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le

1.	The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le (pi)/(omega)` `=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)` The average value of the voltage one cycle isA. `(V_(0))/(sqrt(2))`B. `(V_(0))/(2)`C. zeroD. `(2V_(0))/(pi)`
Answer» Correct Answer - D The average value of the voltage is `V_(av)=(int_(0)^(2pi//omega)Vdt)/(int_(0)^(2pi//omega)dt)=(int_(0)^(pi//omega)V_(0)sin omega dt + int_(pi//omega)^(2pi//omega)(-V_(0)sin omega t)dt)/((2pi)/(omega))` `=(omega)/(2pi)[\|(-V_(0)cos omega t)/(omega)\|_(0)^(pi//omega) + \|(V_(0)cos omega t)/(omega)\|_(pi//omega)^(2pi//omega)]=(2V_(0))/(pi)`

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The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le (pi)/(omega)` `=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)` The average value of the voltage one cycle isA. `(V_(0))/(sqrt(2))`B. `(V_(0))/(2)`C. zeroD. `(2V_(0))/(pi)`

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