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Question: The average value of alternating current over complete cycle is A. Zero B. \(1\text{ }rms\) C....

The average value of alternating current over complete cycle is
A. Zero
B. 1 rms1\text{ }rms
C. I2\dfrac{I}{\sqrt{2}}
D. 2I2I

Explanation

Solution

Hint: Obtain the equation for the alternating current by understanding the concept behind it. We can find the total current by integrating it over the time period of the complete cycle. Then we can find the average current over a complete cycle and divide the quantity obtained by the time period.

Complete step by step answer:
Consider a DC current. It is an electric current that flows in a single direction only. If we consider an Ac current or alternating current, it changes the direction of flow of current periodically. It changes the polarity of current or voltage periodically.
We can define the alternating current in terms of a sinusoidal wave which has the same property as an AC i.e. it also produces waves with alternate polarity.
AC current can be mathematically expressed as,
I=I0sinwtI={{I}_{0}}\sin wt
Where, I0{{I}_{0}} is the maximum current or the peak current.
Let, the time period of one complete cycle is T.
We can write the total current of a full cycle by integrating the current equation over time for the time period of one complete cycle,
Total current =0TI0sinwtdt=\int_{0}^{T}{{{I}_{0}}\sin wtdt}
Average value of alternating current over complete cycle is given as,
=0TI0sinwtdt0Tdt =0TI0sinwtdtT =I0T0Tsinwtdt \begin{aligned} & =\dfrac{\int_{0}^{T}{{{I}_{0}}\sin wtdt}}{\int_{0}^{T}{dt}} \\\ & =\dfrac{\int_{0}^{T}{{{I}_{0}}\sin wtdt}}{T} \\\ & =\dfrac{{{I}_{0}}}{T}\int_{0}^{T}{\sin wtdt} \\\ \end{aligned}
Let, wt=θ where, as t0, θ0 and as tT,θ2π and wdt=dθ \begin{aligned} & wt=\theta \text{ where, as t}\to \text{0, }\theta \to \text{0 and as t}\to T,\theta \to 2\pi \\\ & \text{and wdt=d}\theta \\\ \end{aligned}
So, we can write,
=I0T02πsinθ1wdθ =I0Tw02πsinθdθ =I0Tw[cosθ]02π =I0Tw[cos0cos2π] =I0Tw[11] =0 \begin{aligned} & =\dfrac{{{I}_{0}}}{T}\int_{0}^{2\pi }{\sin \theta }\dfrac{1}{w}d\theta \\\ & =\dfrac{{{I}_{0}}}{Tw}\int_{0}^{2\pi }{\sin \theta d\theta } \\\ & =\dfrac{{{I}_{0}}}{Tw}\left[ -\cos \theta \right]_{0}^{2\pi } \\\ & =\dfrac{{{I}_{0}}}{Tw}\left[ \cos 0-\cos 2\pi \right] \\\ & =\dfrac{{{I}_{0}}}{Tw}\left[ 1-1 \right] \\\ & =0 \\\ \end{aligned}
So, the average value of the alternating current is zero.
The correct option is (A).
Note: In DC current we always have a constant value. So, we don’t have a rms value of average value for DC current. In alternating current, the polarity of the current or the voltage changes periodically and that’s why we get rms values or average values for alternating current.