In india we use alternating voltage with rms value of 220 V and frequency 50 Hz. (a) Calculate corresponding peak voltage . (b) What minimum time the voltage takes to change from 0 to rms value ? (c ) What maximum time will it take to become 0 starting from rms value ?

Explanation

(a) Peak value of voltage is given by
`epsilon_(0)= sqrt(2) epsilon_("rms")`
Here `epsilon_("rms")= 220 V`
`:. " " epsilon_(0)= sqrt(2)xx220 = 1.414xx220 = 311 V `
(b) Let the alternating voltage be given by
`epsilon= epsilon_(0) sin omegat `
Let `t_(1)` be the minimum time taken by the voltage to take its rms value.
Then `epsilon= epsilon_("rms")`
`implies epsilon_("rms")=sqrt(2) epsilon_("rms") sin omegat_(1)`
`implies (1)/(sqrt(2))= sin omegat_(1)`
`implies omegat_(1)= (pi)/(4) `
`implies t_(1) = (pi)/(4)`
`implies t_(1)= (pi)/(4omega)`
Here `omega= 2pif= 2pixx50 " " (because f = 50 Hz)`
`implies t_(1)= (pi)/(4xx2pixx50)=(2.5 ms `
(c ) Let the time taken by voltage to become zero be `t_(2)`.
`0 =epsilon_(0) sin omegat_(2)`
`implies sin omega t_(2)=0`
`implies omegat =pi`
`implies t_(2)= (pi)/(omega)= (pi)/(2pixx50)= 10 ms `
`t_(2)-t_(1)` = 10 ms - 2.5 ms = 7.5 ms
This will be the maximum time taken by voltage to become zero starting from rms value.