The average value of alternating current during a full cycle is (i₀ is the peak value): 

Concept:

Alternating current is defined as the current whose magnitude changes with time and also reverses its direction periodically. The maximum value reached by an alternating quantity in one cycle is known as the peak value.

The average value is defined as the average of all the instantaneous values of an alternating quantity such as current or voltage over one complete cycle.

AC current can be expresses as

• \(I=I_osin\omega t\)

where \(I_o\) is the maximum or 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 \(I=\int _0^TI_osin\omega t.dt\)

Average value of alternating current over complete cycle is given as,

• \(I_{av}=\frac{\int_0^T I_osin\omega t dt}{\int_0^Tdt}\)

Explanation:

\(I_{av}=\frac{\int_0^T I_osin\omega t dt}{\int_0^Tdt}\)

\(I_{av}=\frac{\int_0^T I_osin\omega t dt}T\)

\(I_{av}=\frac{I_o}{T} {\int_0^Tsin\omega t dt}\)

Let \(\omega t=\theta\), where, as \(t\to0,\ \theta\to0\ and\ as\ t\to T,\ \theta\to2\pi\ and\ \omega dt=d\theta\),

So, we can write, 

\(I_{av}=\frac{I_o}{T}\int_0^{2\pi}sin\theta\frac{1}{\omega}.dt\)

\(I_{av}=\frac{I_o}{\omega T}\int_0^{2\pi}sin\theta.dt\)

\(I_{av}=\frac{I_o}{\omega T}[-cos\theta]_0^{2\pi}\)

\(I_{av}=\frac{I_o}{\omega T}[cos0-cos2\pi]\)

\(I_{av}=\frac{I_o}{\omega T}[1-1]\)

\(I_{av}=0\)

Hence, the correct answer is Option-2-zero.