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Alternating Current — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
Define mean value and root mean square value of alternating current.
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Answer

Mean Value : Alternating voltage and alternating current both vary in magnitude with time and change the direction periodically. During one of the half cycles these have positive value whereas in the next half negative value. Thus the mean value of alternating voltage and alternating current each for a complete cycle is zero. Because of this an ordinary voltmeter or ammeter gives no deflection in alternating current circuit.
But for any half cycle the mean value of alternating voltage or alternating current is not zero, which can be shown as follows:
The mean value of alternating current for first positive half cycle is expressed as follows:
$I_{m}=\frac{\int_{0}^{T/2}Idt}{T/2}=\frac{\int_{0}^{T/2}(I_{0}sin \omega t)dt}{T/2}$
or $I_{m}=\frac{\int_{0}^{T/2}I_{0}sin(\frac{2\pi}{T}).tdt}
{T/2} = \frac{2I_{0}}{T}\frac{[-cos\frac{2\pi}{T}.t]_{0}^{T/2}}{\frac{2\pi}{T}}$
or $I_{m}=-\frac{2I_{0}}{T}\times\frac{T}{2\pi}[cos \pi-cos0]=-\frac{I_{0}}{\pi}[-1-1]$
or $I_{m}=+(\frac{2I_{0}}{\pi})$
or $I_{m}=+\frac{2I_{0}}{3.14}\Rightarrow I_{m}=+0.637~I_{0}$
Similarly, for next negative half cycle the average value of alternating current will be :
$I_{m}=-(\frac{2I_{0}}{\pi})$ or $I_{m}=-0.637~I_{0}$
Thus the mean value of alternating current for one complete cycle is:
$I_{m}=(+\frac{2I_{0}}{\pi})+(-\frac{2I_{0}}{\pi})=0$
Root Mean Square Value : Although the mean value of alternating voltage and alternating current both for one complete cycle is zero, the mean value of their square for one complete cycle is not zero. Therefore, rest mean square value of these is very important for alternating current circuits.
Root mean square value of alternating current is defined as follows:
"The square root of the mean value of the square of alternating current for its one complete cycle is called its root mean square value." It is denoted by $I_{rms}$ and is calculated as follows:
Mean value of $I^{2}$ for one complete cycle is given by:
$\overline{ I ^2}=\frac{\int_0^{ T } I ^2 \cdot d t}{T} \Rightarrow \overline{ I ^2}=\frac{\int_0^{ T }\left( I _0 \sin \omega t\right)^2 \cdot d t}{T}$
$\Rightarrow \quad \overline{ I ^2}=\frac{\int_0^{ T } I _0^2 \sin ^2 \omega t \cdot d t}{T}=\frac{\int_0^{ T } I _0^2 \sin ^2\left(\frac{2 \pi}{T}\right) \cdot t \cdot d t}{T}$
$\Rightarrow \quad \overline{ I ^2}=\frac{\int_0^{ T } I _0^2 2 \sin ^2\left(\frac{2 \pi}{T}\right) \cdot t \cdot d t}{2 T}$
or $\overline{ I ^2}=\frac{ I _0^2}{2 T} \int_0^{ T }\left[1-\cos 2 \times\left(\frac{2 \pi}{T}\right) \cdot t\right] d t$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\int_0^{ T } d t-\int_0^{ T } \cos \left(\frac{4 \pi}{T}\right) \cdot t \cdot d t\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\frac{\{t\}_0^{ T }-\left[\sin \left(\frac{4 \pi}{T}\right) \cdot t\right]_0^{ T }}{\frac{4 \pi}{T}}\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\{ T \}-\left\{\frac{ T }{4 \pi}\left(\sin 4 \pi-\sin 0^0\right)\right\}\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T} \times T -\frac{ I _0^2}{2 T} \times 0 \Rightarrow \overline{ I ^2}=\frac{ I _0^2}{2}$
But according to the definition of root mean square value of alternating current it is given by :
$I _{r m s}=\sqrt{\overline{\overline{ I ^2}}} \Rightarrow I _{r m s}=\sqrt{\left(\frac{ I _0^2}{2}\right)}$
or $I _{r m s}=\frac{ I _0}{\sqrt{2}}$
or $I_{r m s}=\frac{I_0}{1.414} \Rightarrow I_{r m s}=0.707 I_0$

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