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  4. A direct current of 5 A is superimposed on an alternating current I=10 sin (ω t) flowing through a wire. The effective value of the resulting current will be

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Q. A direct current of $5 A$ is superimposed on an alternating current $I=10 \sin (\omega t)$ flowing through a wire. The effective value of the resulting current will be

Alternating Current

Solution:

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