If `i_(1)=3 sin omega t and (i_2) = 4 cos omega t,` then `(i_3)` is

Answer

`5 sin(omega t+ 53^(0))`

Explanation

`i_(3)=i_(1)+i_(2)`
`=3 sin omega t + 4 sin (omega t +(pi)/(2))`
`= 5 sin (omega t + "tan"^(-1)(4)/(3))`