Question

An alternating current is given by i = i₁ cos ωt + i₂ sin ωt. The rms current is given by

(a) $\frac{i_1+i_2}{\sqrt{2}}$
(b) $\frac{\left|i_1+i_2\right|}{\sqrt{2}}$
(c) $\sqrt{\frac{i_1^2+i_2^2}{2}}$
(d) $\sqrt{\frac{i_1^2+i_2^2}{\sqrt{2}}}$

Answer 1

(c) $\sqrt{\frac{i_1^2+i_2^2}{2}}$
Given:
i = i₁ cos ωt + i₂ sin ωt
The rms value of current is given by,
$$\begin{gathered} i_{rms}=\sqrt{\frac{{\int }_0^T{i}^{2}dt}{{\int }_0^{T}dt}} \\ i=i_1\cos \omega t+i_2\sin \omega t \\ {i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_0^T{\left(i_1\cos \omega t+i_2\sin \omega t\right)}^{2}dt}{{\int }_0^{T}dt}} \\ {i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_0^T\left(i_1^2{\cos }^2\omega t+i_2^2{\sin }^2\omega t+2i_1{i}_2\sin \mathrm{\omega t}\cos \omega t\right)dt}{{\int }_0^{T}dt}} \\ {i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_0^T\left(i_1^2{\displaystyle \frac{(\cos 2\omega t+1)}{2}}+i_2^2{\displaystyle \frac{(1-\cos 2\omega t)}{2}}+i_1{i}_2\sin 2\omega t\right)dt}{{\int }_0^{T}dt}} \\ [\because {\cos }^2\omega t=\frac{(\cos 2\omega t+1)}{2},{\sin }^2\omega t=\frac{(1-\cos 2\omega t)}{2}] \end{gathered}$$

We know that, T = 2π

Integrating the above expression
$i_{\mathrm{rms}}=\sqrt{\frac{{\displaystyle \frac{1}{2}}{i}_1^2\left({\int }_0^{2\pi }1dt+{\int }_0^{2\pi }\cos 2\omega tdt\right)+i_2^2\left({\int }_0^{2\pi }1dt-{\int }_0^{2\pi }\cos 2\omega tdt\right)+i_1{i}_2{\int }_0^{2\pi }\sin 2\omega tdt}{{\int }_0^{2\pi }dt}}$

The following integrals become zero
${\int }_0^{2\pi }\cos 2\omega tdt=0={\int }_0^{2\pi }\sin 2\omega tdt$
Therefore, it becomes
$$\begin{gathered} i_{\mathrm{rms}}=\sqrt{\frac{{\displaystyle \frac{i_1^2}{2}}\left({\int }_0^{2\mathrm{\pi }}1dt\right)+{\displaystyle \frac{i_2^2}{2}}\left({\int }_0^{2\mathrm{\pi }}1dt\right)}{{\int }_0^{2\pi }dt}} \\ {i}_{\mathrm{rms}}=\sqrt{\frac{{\displaystyle \frac{i_1^2}{2}}\times 2\mathrm{\pi }+{\displaystyle \frac{i_2^2}{2}}\times 2\mathrm{\pi }}{2\mathrm{\pi }}} \\ \Rightarrow i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}} \end{gathered}$$