Question

The root mean square value of current is \[{{I}_{rms}}=\sqrt{\dfrac{\int{{{I}^{2}}dt}}{\int{dt}}}\] Current through a wire is \[I={{I}_{0}}\sin \omega t\]. Find the root mean square value of current from \[t=0\] to \[t=\dfrac{2\pi }{\omega }\] .
\[\begin{align}
  & A){{I}_{0}} \\
 & B){{I}_{0}}^{2} \\
 & C)\dfrac{2{{I}_{0}}}{\pi } \\
 & D)\dfrac{{{I}_{0}}}{\sqrt{2}} \\
\end{align}\]

Answer 1

Hint: Here, current through the circuit and root mean square value of current are given. To find the root means square value at a given time period, substitute the value of current through the circuit in rms equation of current and upon integrating it, we can determine the value of \[{{I}_{rms}}\]at that particular time interval.

Formula used:
\[{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}\]

Complete step by step solution:
Given,
Current through a wire , \[I={{I}_{0}}\sin \omega t\] ---------- (1)
The root mean square value of current,\[{{I}_{rms}}=\sqrt{\dfrac{\int{{{I}^{2}}dt}}{\int{dt}}}\] --------- (2)
Substitute 1 in equation 2, we get,
\[{{I}_{rms}}=\sqrt{\dfrac{\int{{{\left( {{I}_{0}}\sin \omega t \right)}^{2}}dt}}{\int{dt}}}\]
Squaring the above equation,
\[I_{rms}^{2}=\dfrac{\int{{{\left( {{I}_{0}}\sin \omega t \right)}^{2}}dt}}{\int{dt}}=\dfrac{\int{\left( {{I}_{0}}^{2}{{\sin }^{2}}\omega t \right)dt}}{\int{dt}}\] ----------- (3)
We have,
\[{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}\]
Then, equation 3 becomes,
\[I_{rms}^{2}=\dfrac{\int{{{I}_{0}}^{2}\left( \dfrac{1-\cos 2\omega t}{2} \right)dt}}{\int{dt}}=\dfrac{{{I}_{0}}^{2}\int{\left( \dfrac{1-\cos 2\omega t}{2} \right)dt}}{\int{dt}}\]
Integrating from\[t=0\] to \[t=\dfrac{2\pi }{\omega }\] ,
\[I_{rms}^{2}=\dfrac{{{I}_{0}}^{2}\int\limits_{0}^{\dfrac{2\pi }{\omega }}{\left( \dfrac{1-\cos 2\omega t}{2} \right)dt}}{\int\limits_{0}^{\dfrac{2\pi }{\omega }}{dt}}=\dfrac{{{I}_{0}}^{2}\left( \dfrac{1}{2}-\dfrac{\sin 2\omega t}{2} \right)_{0}^{\dfrac{2\pi }{\omega }}}{\left( t \right)_{0}^{\dfrac{2\pi }{\omega }}}=\dfrac{{{I}_{0}}^{2}\left( \dfrac{2\pi }{2\omega } \right)}{\dfrac{2\pi }{\omega }}\]
Then,
\[I_{rms}^{2}=\dfrac{{{I}_{0}}^{2}}{2}\Rightarrow I=\dfrac{{{I}_{0}}}{\sqrt{2}}\]

Therefore, the correct option is option (D).

Additional information:
The measure of the magnitude of a varying quantity is called the root mean square (RMS) value. We use the root mean square to represent the average current or voltage in an AC circuit. The RMS voltage and current (for sinusoidal systems) are the peak voltage and current over the square root of two. The average power in an AC circuit is the product of the RMS voltage and RMS current. DC and AC waveforms can both represent current or voltage waveforms, but they are in different forms.

Note:
 AC waveforms fluctuate between positive and negative cycles. DC voltage is just constant value. Because of this difference, it's difficult to compare the two. But, RMS value gives us a standard to compare the amount of power that a DC waveform and an AC waveform can give to a circuit.