1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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) $\sqrt{\frac{{i}_{1}^{2}+{i}_{2}^{2}}{\sqrt{2}}}$

Open in App
Solution

## (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, ${i}_{rms}=\sqrt{\frac{{\int }_{0}^{T}{i}^{2}dt}{{\int }_{0}^{T}dt}}\phantom{\rule{0ex}{0ex}}i={i}_{1}\mathrm{cos}\omega t+{i}_{2}\mathrm{sin}\omega t\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_{0}^{T}{\left({i}_{1}\mathrm{cos}\omega t+{i}_{2}\mathrm{sin}\omega t\right)}^{2}dt}{{\int }_{0}^{T}dt}}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_{0}^{T}\left({i}_{1}^{2}{\mathrm{cos}}^{2}\omega t+{i}_{2}^{2}{\mathrm{sin}}^{2}\omega t+2{i}_{1}{i}_{2}\mathrm{sin}\mathrm{\omega t}\mathrm{cos}\omega t\right)dt}{{\int }_{0}^{T}dt}}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{rms}}=\sqrt{\frac{{\int }_{0}^{T}\left({i}_{1}^{2}\frac{\left(\mathrm{cos}2\omega t+1\right)}{2}+{i}_{2}^{2}\frac{\left(1-\mathrm{cos}2\omega t\right)}{2}+{i}_{1}{i}_{2}\mathrm{sin}2\omega t\right)dt}{{\int }_{0}^{T}dt}}\phantom{\rule{0ex}{0ex}}\left[\because {\mathrm{cos}}^{2}\omega t=\frac{\left(\mathrm{cos}2\omega t+1\right)}{2},{\mathrm{sin}}^{2}\omega t=\frac{\left(1-\mathrm{cos}2\omega t\right)}{2}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ We know that, T = 2π Integrating the above expression ${i}_{\mathrm{rms}}=\sqrt{\frac{\frac{1}{2}{i}_{1}^{2}\left({\int }_{0}^{2\pi }1dt+{\int }_{0}^{2\pi }\mathrm{cos}2\omega tdt\right)+{i}_{2}^{2}\left({\int }_{0}^{2\pi }1dt-{\int }_{0}^{2\pi }\mathrm{cos}2\omega tdt\right)+{i}_{1}{i}_{2}{\int }_{0}^{2\pi }\mathrm{sin}2\omega tdt}{{\int }_{0}^{2\pi }dt}}$ The following integrals become zero ${\int }_{0}^{2\pi }\mathrm{cos}2\omega tdt=0={\int }_{0}^{2\pi }\mathrm{sin}2\omega tdt\phantom{\rule{0ex}{0ex}}$ Therefore, it becomes ${i}_{\mathrm{rms}}=\sqrt{\frac{\frac{{i}_{1}^{2}}{2}\left({\int }_{0}^{2\mathrm{\pi }}1dt\right)+\frac{{i}_{2}^{2}}{2}\left({\int }_{0}^{2\mathrm{\pi }}1dt\right)}{{\int }_{0}^{2\pi }dt}}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{rms}}=\sqrt{\frac{\frac{{i}_{1}^{2}}{2}×2\mathrm{\pi }+\frac{{i}_{2}^{2}}{2}×2\mathrm{\pi }}{2\mathrm{\pi }}}\phantom{\rule{0ex}{0ex}}⇒{i}_{\mathrm{rms}}=\sqrt{\frac{{i}_{1}^{2}+{i}_{2}^{2}}{2}}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Trigonometric Ratios of Allied Angles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program