Question

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

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

Answer 1

Answer: (B)

Given: $I=5+10\sin \omega t$
$I_{eff}{\left[\frac{\underset{0}{\overset{T}{\int }}{I}^{2}dt}{\underset{0}{\overset{T}{\int }}dt}\right]}^{1/2}$
$={\left[\frac{1}{T}\underset{0}{\overset{T}{\int }}(5+10\sin \omega t{)}^{2}dt\right]}^{1/2}$
$=[\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left(25+100\sin \omega t+100{\sin }^2\omega t\right]}^{1/2}$
But as $\frac{1}{T}\underset{0}{\overset{T}{\int }}\sin \omega tdt=0$
and $\frac{1}{T}\underset{0}{\overset{T}{\int }}{\sin }^2\omega tdt=\frac{1}{2}$
so $I_{eff}={\left[25+\frac{1}{2}\times 100\right]}^{1/2}$
$=5\sqrt{3}A$