An alternating voltage is given by `E = 100 sin( omega +(pi)/(6))V`. The voltage will be maximum for the first time when is [T = periodic time)

Options

A. `(T)/(12)`
B. `(T)/(2)`
C. `(T)/(6)`
D. `(T)/(3)`

Answer

C

Explanation

Given , alternating voltage is
` E = 100 sin ( omega t + ( pi ) /( 6 ) ) V ` ltBrgt The voltage will maximum, when
`sin ( omega t + ( pi ) / ( 6 ) ) = 1 `
` rArr sin ( omega t+ ( pi ) / ( 6 ) ) = sin"" (pi ) /(6) `
` rArr omega t+ ( pi ) / ( 6 ) = ( pi ) / ( 2 ) `
` rArr omega t = ( pi ) / ( 3 ) `
` rArr t = ( pi ) /( 3 omega ) = ( pi xx T ) /( 3 xx 2pi ) " " [ because omega = ( 2 pi ) /( T) ] `
` = ( T ) /( 6 ) `
Thus, the voltage will be maximum for the first time when ` t = ( T ) / ( 6 ) `.