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

Given $I=5+10\sin \omega t$ 

$I_{eff}={\left[\frac{{\displaystyle \underset{0}{\overset{T}{\int }}{I}^{2}dt}}{{\displaystyle \underset{0}{\overset{T}{\int }}dt}}\right]}^{1/2}$

$={\left[\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}{\left(5+10\sin \omega t\right)}^{2}dt}\right]}^{1/2}$

$={\left[\frac{1}{T}\underset{0}{\overset{T}{\int }}(25+100\sin \omega t+100{\sin }^2\omega t)dt\right]}^{1/2}$

But as $\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}\sin \omega tdt=0}$

And $\frac{1}{T}\underset{0}{\overset{T}{\int }}{\sin }^2\omega tdt=\frac{1}{2}$

So $I_{eff}={[25+\frac{1}{2}\times 100]}^{1/2}=5\sqrt{3}A$