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A direct current of $6 \mathrm{~A}$ is superimposed on an alternating current $I=10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be
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Verified Answer
The correct answer is:
$9.27$
$9.27$
Given, $I=6+10 \sin \omega t$
$$
\begin{aligned}
I_{\mathrm{eft}} & =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omega t)^2 d t\right]^{1 / 2}\right. \\
& =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omega t+100 \sin ^2 \omega t\right) d t\right]^{1 / 2}
\end{aligned}
$$
But as, $\frac{1}{T} \int_0^T \sin \omega t d t=0$ and $\frac{1}{T} \int_0^T \sin ^2 \omega t=\frac{1}{2}$
$$
\Rightarrow \quad I_{\mathrm{eff}}=\left[36+\frac{1}{2} \times 100\right]^{1 / 2}=9.27
$$
Thus, $\quad I_{\text {eff }}=9.27 \mathrm{~A}$.
$$
\begin{aligned}
I_{\mathrm{eft}} & =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omega t)^2 d t\right]^{1 / 2}\right. \\
& =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omega t+100 \sin ^2 \omega t\right) d t\right]^{1 / 2}
\end{aligned}
$$
But as, $\frac{1}{T} \int_0^T \sin \omega t d t=0$ and $\frac{1}{T} \int_0^T \sin ^2 \omega t=\frac{1}{2}$
$$
\Rightarrow \quad I_{\mathrm{eff}}=\left[36+\frac{1}{2} \times 100\right]^{1 / 2}=9.27
$$
Thus, $\quad I_{\text {eff }}=9.27 \mathrm{~A}$.
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