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. `(15//2)A`
B. `5 sqrt(3)A`
C. `5 sqrt(5)A`
D. `15 A`

Answer

B

Explanation

Given `l=5+10 sin omega t `
`I_(eff)=[(int_(0)^(T)(i^2)dt)/(int_(0)^(T)dt)] = [ 1/T int_(0)^(T) (5 + 10 sin omega t)^(2) dt ]^(1//2)`
`[ 1/T int_(0)^(T) (25 + 100 sin omega t)+100 sin^(2)omega t)]^(1//2)`
But as `(1)/(T) int_(0)^(T) sin omega t dt = 0`
and ` 1/T int_(0)^(T) sin^(2) omega t dt = 1/2`
so `I_(eff) = [ 25 + 1/2 xx 100 ]^(1//2) = 5 sqrt(3)A`.