Question

An alternative current is given by $i=i_1\cos \omega t+i_2\sin \omega t$. The $rms$ current is given by

• A. $\frac{i_1+i_2}{\sqrt{2}}$
• B. $\frac{|i_1+i_2|}{\sqrt{2}}$
• C. $\sqrt{\frac{i_1^2+i_2^2}{2}}$
• D. $\sqrt{\frac{i_1+i_2}{\sqrt{2}}}$

Answer 1

Answer: (C)

$\sqrt{\frac{i_1^2+i_2^2}{2}}$

Answer: C $\sqrt{\frac{i_1+i_2}{2}}$
Time period of $i$ is $\tau =\frac{2\pi }{\omega }$
$i_{rms}^2=\frac{1}{\tau }{\int }_0^{\tau }{i}^{2}dt=\frac{1}{\tau }{\int }_0^{\tau }{\left(i_{1}cos\right(\omega t)+i_2\sin (\omega t\left)\right)}^{2}dt=\frac{1}{\tau }{\int }_0^{\tau }\left(i_1^2{\cos }^2\right(\omega t)+i_2^2{\sin }^2(\omega t)+2i_2{i}_2\cos (\omega t\left)\sin \right(\omega t\left)\right)dt=\frac{1}{\tau }\left[{\int }_0^{\tau }{i}_1^2{\cos }^2\left(\omega t\right)dt+{\int }_0^{\tau }{i}_2^2{\sin }^2\left(\omega t\right)dt+{\int }_0^{\tau }{i}_1{i}_{2}sin\left(2\omega t\right)dt\right]=\frac{1}{\tau }[\frac{1}{2}{i}_1^2\tau +\frac{1}{2}{i}_2^2\tau +0]=\frac{1}{2}[i_1^2+i_2^2]$
Therefore,
$i_{rms}=\sqrt{\frac{i_1^2+i_2^2}{2}}$