Question

The rms value of current $I_{rms}$ is (where, $I_{0}$ is the value of peak current)
\[\begin{align}
  & A.\text{ }\dfrac{{{I}_{0}}}{2\pi } \\
 & B.\text{ }\dfrac{{{I}_{0}}}{\sqrt{2}} \\
 & C.\text{ }\dfrac{2{{I}_{0}}}{\pi } \\
 & ~D.\text{ }\sqrt{2}{{I}_{0}} \\
\end{align}\]

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, $dH=(I_{0}\sin \omega t)^{2}Rdt$.

Complete step-by-step solution:
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 the fill 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.
Consider $I=I_{0}\sin\omega t$, then the heat produced $dH=I^{2}Rdt$
$dH=(I_{0}\sin \omega t)^{2}Rdt$
Then the heat produced in half period is,
\[\begin{align}
  & H=\int\limits_{0}^{\dfrac{T}{2}}{I_{0}^{2}R{{\sin }^{2}}\omega t}dt \\
 & =I_{0}^{2}R\int\limits_{0}^{\dfrac{T}{2}}{{{\sin }^{2}}\omega t}dt \\
 & =I_{0}^{2}R\int\limits_{0}^{\dfrac{T}{2}}{\dfrac{1}{2}[1-\cos (2\omega t)]}dt \\
 & =\dfrac{I_{0}^{2}R}{2}\left[ t-0 \right]_{0}^{\dfrac{T}{2}} \\
 & =\dfrac{I_{0}^{2}R}{2}\left[ \dfrac{T}{2}-0 \right]=\dfrac{I_{0}^{2}RT}{4} \\
\end{align}\]
The rms of alternating current is represented as \[H={{I}^{2}}_{rms}R\dfrac{T}{2}\]
Then equating, we get
\[{{I}^{2}}_{rms}R\dfrac{T}{2}=\dfrac{I_{0}^{2}RT}{4}\]
Simplifying, we get
\[{{I}^{2}}_{rms}=\dfrac{I_{0}^{2}}{2}\]
Thus, the rms value of current $I_{rms}$ is \[{{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}\]
Hence the answer is B. $\dfrac{I_{0}}{\sqrt 2}$

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 \[{{I}_{rms}}=\dfrac{{{I}_{0}}}{\sqrt{2}}=0.707{{I}_{0}}\]