A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:

Options

A. `7.5A`
B. `2sqrt3A`
C. `5sqrt3A`
D. `15A`

Answer

C

Explanation

`I =5 + 10 sin omegat`
`bari^(2) = 5^(2) + 10^(2)/(2) =75`
`i_(rms) = sqrt(bari^(2)) = 5 sqrt3 A` .