Current in an `AC` circuit is given by `i = 3 sin omegat + 4 cos omegat`, then

Options

A. `(I_(1)+I_(2))/(2)`
B. `((I_(1)+I_(2)^(2))/(sqrt(2)))`
C. `sqrt((I_(1)^(2)+I_(2)^(2))/(2))`
D. `(I_(1)^(2)+I_(2)^(2))/(2)`

Answer

C

Explanation

`I=(I_1) cos omega t +(I_2) sin omega t`
`(I^(2))_(mean)=I_(1)^(2) bar(cos^(2)omega t) + I_(2)^(2) bar(sin^(2) omega t) + I_(1)I_(2) bar(cos omega t sin omega t)`
`I_(1)^(2) xx 1/2 xx i_(2)^(2) xx 1/2 + 2I_(1)I_(2) xx 0`
`I_(rms)=sqrt((i^(2))_(mean))=sqrt((I_(1)^(2)+I_(2)^(2))/(2))`.