Qstnid -Ii-9-an Alternating Current Is Given By I = I<sub>1</sub> Cos Ωt + I<sub>2</sub> Sin Ωt. The Rms Current Is Given By - 39: Alternating Current

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NEET-XII-Physics

39: Alternating Current

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Qstn# ii-9 Prvs-Qstn Next-Qstn

#9
An alternating current is given by i = i₁ cos ωt + i₂ sin ωt. The rms current is given by
(a)
i1+i22
(b)
i1+i22
(c)
i12+i222
(d)
i12+i222
digAnsr: c
Ans : (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}}``
`` i={i}_{1}\,\mathrm{\,cos\,}\omega t+{i}_{2}\,\mathrm{\,sin\,}\omega t``
`` ``
`` {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}}``
`` {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}}``
`` {i}_{\,\mathrm{\,rms\,}}=\sqrt{\frac{{\int }_{0}^{T}\left({i}_{1}^{2}{\displaystyle \frac{(\,\mathrm{\,cos\,}2\omega t+1)}{2}}+{i}_{2}^{2}{\displaystyle \frac{(1-\,\mathrm{\,cos\,}2\omega t)}{2}}+{i}_{1}{i}_{2}\,\mathrm{\,sin\,}2\omega t\right)dt}{{\int }_{0}^{T}dt}}``
`` [\because {\,\mathrm{\,cos\,}}^{2}\omega t=\frac{(\,\mathrm{\,cos\,}2\omega t+1)}{2},{\,\mathrm{\,sin\,}}^{2}\omega t=\frac{(1-\,\mathrm{\,cos\,}2\omega t)}{2}]``
`` ``
`` ``
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 }\,\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``
`` ``
Therefore, it becomes
`` {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}}``
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