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:
\[\begin{align}
  & A.2 \\
 & B.\sqrt{2} \\
 & C.2\sqrt{2} \\
 & D.0 \\
\end{align}\]

Answer 1

Hint: 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,
\[\begin{align}
  & H=\int\limits_{0}^{\dfrac{T}{2}}{{{2}^{2}}R{{\cos }^{2}}(\omega t}+\theta )dt \\
 & =4R\int\limits_{0}^{\dfrac{T}{2}}{{{\cos }^{2}}(\omega t+\theta )}dt \\
 & =4R\int\limits_{0}^{\dfrac{T}{2}}{\dfrac{1}{2}[1+\cos (2\omega t)]}dt \\
 & =\dfrac{4R}{2}\left[ T-0 \right]_{0}^{\dfrac{T}{2}} \\
 & =\dfrac{I_{0}^{2}R}{2}\left[ \dfrac{T}{2}-0 \right]=\dfrac{4RT}{4} \\
\end{align}\]
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 \[{{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}=0.707{{I}_{0}}\]