Two alternating voltage generators produce emfs of the same amplitude`(E_0)` but with a phase difference of `(pi)//3`. The resultant emf is

Options

A. (A) `E_0 sin [ omega t+(pi)//3)]`
B. (B) `E_0 sin [ omega t+(pi)//6)]`
C. (C) `sqrt(3)E_0 sin [omega t+(pi)//6)]`
D. (D) `sqrt(3)E_0 sin [ omega t+(pi)//2)]`

Answer

C

Explanation

`E_(1)=(E_0) sin omega t, (E_2)=(E_0) sin [ omega t + (pi//3)]`
`E=(E_2)+(E_1)`
`=(E_0)sin (omega t +(pi//3)]+(E_0)sin omegat`
`(E_0) [ 2 sin {omega t+(pi//6)}cos (pi//6)]`
`= sqrt(3) (E_0) sin [omega t +(pi//6)]`.