Question

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

Answer 1

Correct Answer - B
Given `1=5+10 sin omegat`
`I_(eff)=[(int_(0)^(T)I^(2)dt)/(int_(0)^(T) dt)]^(1//2)`
`=[1/Tint_(0)^(T)(5+10 sin omega t)^(2) dt]^(1//2)`
`=[1/Tint_(0)^(T) (25+100 sin omegat+100 sin^(2) omegat)]^(1//2)`
But as `1/Tint_(0)^(T) sin omega t dt=0`
and `1/Tint_(0)^(T) sin^(2) omega t dt=1/2`
so `I_(eff)=[25+1/2xx100]^(1//2)`
`=5sqrt(3)` amp