Question

A direct current of $5A$ is superposed on an alternative current $I=10\sin \omega t$ flowing through the wire. The effective value of the resulting current will be

• A. $\left(15/2\right)A$
• B. $5\sqrt{3}A$
• C. $5\sqrt{5}A$
• D. $15A$

Answer 1

The correct option is B $5\sqrt{3}A$

Total current, $I=(5+10\sin \omega t)$
$\Rightarrow I_{eff}={\left[\frac{{\int }_0^T{I}^{2}dt}{{\int }_0^{T}dt}\right]}^{1/2}$
$={\left[\frac{1}{T}{\int }_0^T\right(5+10\sin \omega t{)}^{2}dt]}^{1/2}$
$={\left[\frac{1}{T}{\int }_0^T\right(25+100\sin \omega t+100{\sin }^2\omega t\left)\right]}^{1/2}$
But, $\frac{1}{T}{\int }_0^T\sin \omega t.dt=0$ and $\frac{1}{R}{\int }_0^1{\sin }^2\omega t.dt=\frac{1}{2}$
So, $I_{eff}={[25+\frac{1}{2}\times 100]}^{1/2}=5\sqrt{3}A$.