A direct current of 5A 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)`
C. `5sqrt(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//2) = [ (1)/(T) int_(0)^(T) ( 5 + 10 sin omegat)^(2)dt]^(1//2)`
` = [(1)/(T) int_(0)^(T) (25+ 100 sin omegat dt =0 and (1)/(T)int_(0)^(T) sin^(2) omegat dt =(1)/(2)`
So `I_("eff")=[25+1/2xx100]^(1//2) = 5sqrt(3)A`