A current is made up of two components 3 A de component and ac component given by `I = 4 sin (omega)t A`. The effective value of current is

Options

A. `I=(3+5sinomegat)A and I_(rms)=sqrt(17)A)`
B. `I=(3+4sinomegat)A and I_(rms)=sqrt(15)A)`
C. `I=(4+4sinomegat)A and I_(rms)=sqrt(17)A)`
D. `I=(3+4sinomegat)A and I_(rms)=sqrt(17)A)`

Answer

D

Explanation

`I=I_(ac)+I_(dc)=(4sinomegat+3)A`
`I_(rms)=sqrt(overline(I^(2)))`
`=sqrt(overline((4sinomegat+3)^(2)))`
`=sqrt(16overline(sin^(2)omegat)+24overline(sinomegat)+9)`
`=sqrt(16xx(1)/(2)+0.9)=sqrt(17)A`