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

Options

A. `5 sin(omega t +53^(@))`
B. `5 sin(omega t +37^(@))`
C. `5 sin(omega t +45^(@))`
D. `5 sin(omega t +35^(@))`

Answer

A

Explanation

`(i_3)=(i_1)+(i_2)=3 sin omega t + 4 cos omega t `
`=3 sin omega t + 4 sin (omega t + 90^(@))`
`(i_(30))=sqrt(I_(10)^(2)+I_(20)^(2))=sqrt(3^(2)+4^(2))=5`
`tan(phi) = 4/3 implies phi = 53^(@)`
so, `(i_3)=(I_(30) sin(omega t+phi)=5 sin (omega t+53^(@)).