Question

If the instantaneous value of current is $I=2 cos(\omega t+\theta)$ ampere in an AC circuit, the rms value of the current in ampere will be:

& A.2 \\\ & B.\sqrt{2} \\\ & C.2\sqrt{2} \\\ & D.0 \\\ \end{aligned}$$

Answer 1

We know that, $I=I_{0}\sin\omega t$ or $I=I_{0}\cos \omega t$ , where $I_{0}$ is the peak value of the alternating current. The RMS or the root-mean-square of instantaneous current is the alternating current given by the direct current through the resistance. It is the area covered in a half cycle. It is the heat produce over half cycle, $dH=(I_{0}\sin \omega t)^{2}Rdt$

Formula used:
${{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}=0.707{{I}_{0}}$

Complete step-by-step answer:
Alternating current is the current whose magnitude varies with time and reverse it direction periodically i.e. after half time period. The general equation is given as: $I=I_{0}\sin\omega t$ or $I=I_{0}\cos \omega t$ , where $I_{0}$ is the peak value of the alternating current.
Since the mean value of alternating current is $0$ for full cycle, due to the symmetry of the sinusoidal wave, we usually calculate the value for half-cycle, only.
The RMS or the root-mean-square of instantaneous current is the alternating current given by the direct current through the resistance. It is the area covered in a half cycle.
Here, given that, $I=2 cos(\omega t+\theta)$ , then, $I_{0}=2$ . Consider, the heat produced $dH=I^{2}Rdt$ .
Then the heat produced in half period is,

& H=\int\limits_{0}^{\dfrac{T}{2}}{{{2}^{2}}R{{\cos }^{2}}(\omega t}+\theta )dt \\\ & =4R\int\limits_{0}^{\dfrac{T}{2}}{{{\cos }^{2}}(\omega t+\theta )}dt \\\ & =4R\int\limits_{0}^{\dfrac{T}{2}}{\dfrac{1}{2}[1+\cos (2\omega t)]}dt \\\ & =\dfrac{4R}{2}\left[ T-0 \right]_{0}^{\dfrac{T}{2}} \\\ & =\dfrac{I_{0}^{2}R}{2}\left[ \dfrac{T}{2}-0 \right]=\dfrac{4RT}{4} \\\ \end{aligned}$$ The rms of alternating current is represented as $$H={{I}^{2}}_{rms}R\dfrac{T}{2}$$ Then equating, we get $${{I}^{2}}_{rms}R\dfrac{T}{2}=\dfrac{4RT}{4}$$ Simplifying, we get $${{I}^{2}}_{rms}=\dfrac{4}{2}=2$$ Thus, the rms value of current $ I_{rms} $ is $${{I}_{rms}}=\sqrt{2}$$ Hence the answer is $$B.\sqrt{2}$$ **So, the correct answer is “Option B”.** **Note:** Since alternating current is periodical and sinusoidal wave, i.e. $ I=I_{0}\sin\omega t $ or $ I=I_{0}\cos \omega t $ it is symmetrical. Hence the current when taken over the time-period is $ 0 $ . Thus we can calculate the wave over the half-time period. Also note that $ dH=I^{2}Rdt $. Thus, The rms value of current $ I_{rms} $ is $${{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}=0.707{{I}_{0}}$$