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An alternating current is given by $i=i_1 \sin \omega t+i_2 \cos \omega t$. The rms current is given by
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Verified Answer
The correct answer is:
$\sqrt{\frac{i_1^2+i_2^2}{2}}$
Equation of alternating current,
$$
\begin{aligned}
& i=i_1 \sin \omega t+i_2 \cos \omega t \\
& \text { Let } i_1=I \cos \phi \\
& \text { and } i=I \sin \phi \\
& i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \\
& =I(\sin \omega t \cos \phi+\cos \omega t \sin \phi) \\
& \Rightarrow \quad i=I \sin (\omega t+\phi) \\
& \therefore \text { rms value of } i_1 \text {, } \\
& i_{\mathrm{rms}}=\frac{I}{\sqrt{2}} \\
&
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{gathered}
i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \\
=I^2 \\
\Rightarrow \quad \sqrt{i_1^2+i_2^2}=I
\end{gathered}
$$
$$
i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right)
$$
From Eqs. (iii) and (iv), we get
$$
i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}}
$$
$$
\begin{aligned}
& i=i_1 \sin \omega t+i_2 \cos \omega t \\
& \text { Let } i_1=I \cos \phi \\
& \text { and } i=I \sin \phi \\
& i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \\
& =I(\sin \omega t \cos \phi+\cos \omega t \sin \phi) \\
& \Rightarrow \quad i=I \sin (\omega t+\phi) \\
& \therefore \text { rms value of } i_1 \text {, } \\
& i_{\mathrm{rms}}=\frac{I}{\sqrt{2}} \\
&
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{gathered}
i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \\
=I^2 \\
\Rightarrow \quad \sqrt{i_1^2+i_2^2}=I
\end{gathered}
$$
$$
i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right)
$$
From Eqs. (iii) and (iv), we get
$$
i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}}
$$
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