Question

The rms value of current $I_{rms}$ is (where, $I_0$ is the value of peak current)
$$\begin{array}{rl}& A.{\displaystyle \frac{I_0}{2\pi }}\\ & B.{\displaystyle \frac{I_0}{\sqrt{2}}}\\ & C.{\displaystyle \frac{2I_0}{\pi }}\\ & D.\sqrt{2}{I}_0\end{array}$$

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,
$$\begin{array}{rl}& H=\underset{0}{\overset{{\displaystyle \frac{T}{2}}}{\int }}{I}_0^{2}R{\sin }^2\omega tdt\\ & =I_0^{2}R\underset{0}{\overset{{\displaystyle \frac{T}{2}}}{\int }}{\sin }^2\omega tdt\\ & =I_0^{2}R\underset{0}{\overset{{\displaystyle \frac{T}{2}}}{\int }}{\displaystyle \frac{1}{2}}[1-\cos (2\omega t)]dt\\ & ={\displaystyle \frac{I_0^{2}R}{2}}{[t-0]}_0^{{\displaystyle \frac{T}{2}}}\\ & ={\displaystyle \frac{I_0^{2}R}{2}}[{\displaystyle \frac{T}{2}}-0]={\displaystyle \frac{I_0^{2}RT}{4}}\end{array}$$
The rms of alternating current is represented as $$H={I^2}_{rms}R{\displaystyle \frac{T}{2}}$$
Then equating, we get
$${I^2}_{rms}R{\displaystyle \frac{T}{2}}={\displaystyle \frac{I_0^{2}RT}{4}}$$
Simplifying, we get
$${I^2}_{rms}={\displaystyle \frac{I_0^2}{2}}$$
Thus, the rms value of current $I_{rms}$ is $$I_{rms}={\displaystyle \frac{I_0}{\sqrt{2}}}$$
Hence the answer is B. ${\displaystyle \frac{I_0}{\sqrt{2}}}$

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 $$I_{rms}={\displaystyle \frac{I_0}{\sqrt{2}}}=0.707I_0$$