The magnitude and direction of alternating current continuosly vary with time. The alternating current is positive during the positive half cycle $\left(0\text{ to }\frac{T}{2}\right)$ and negative during the negative half cycle $\left(\frac{T}{2}\text{ to }0\right)$. Hence, the average or mean value of ac over one complete cycle is zero. However, we can find the mean value over any half cycle by calculating the steady current which sends same amount of charge in a certain time as sent by an ac during the same time.

**Average or mean value of alternating current** is defined as the value of steady current which transfers the same amount of charge through a circuit in a certain time interval as transferred by an alternating current through the same circuit during the same time interval. It is represented by $I_m$.

If $\,i_1,i_2,…,i_n\,$ are the instantaneous currents, then their arithmetic mean gives the mean value of ac. \[I_m=\frac{i_1+i_2+…+i_n}{n}\]

## Relation between Peak Value and Mean Value of AC

The instantaneous alternating current at an instant $t$ is given by \[I=I_0\sin\omega t\text{ __(1)}\]

The small amount of charge $dq$ sent by the alternating current $I$ for small amount of time $dt$ is given by,

\[dq=Idt\]

\[\therefore dq=I_0\sin\omega t\text{ }dt\text{ __(2)}\]

The total charge transferred by ac in the first half cycle can be obtained by integrating above equation from $t=0$ to $t=\frac{T}{2}$.

\[\therefore q=\int_0^{\frac{T}{2}}I_0\sin\omega t\text{ }dt\]

\[=-\frac{I_0}{\omega}\left[\cos\omega t\right]_0^{\frac{T}{2}}\]

\[=-\frac{I_0}{2π/T}\left[\cos\frac{2π}{T}t\right]_0^{\frac{T}{2}}\]

\[=-\frac{I_0T}{2π}\left[\cos\frac{2π}{T}.\frac{T}{2}-\cos 0°\right]\]

\[=-\frac{I_0T}{2π}[\cos π-\cos 0°]\] \[=-\frac{I_0T}{2π}[-1-1]\]

\[\therefore q=\frac{I_0T}{π}\text{ __(3)}\]

If $I_m$ represents the mean value of an ac over a positive half cycle, then the charge sent by it in time $\frac{T}{2}$ is given by,

\[q=I_m\frac{T}{2}\text{ __(4)}\]

From equations $\text{(3)}$ and $\text{(4)}$,

\[I_m\frac{T}{2}=\frac{I_0T}{π}\]

\[\therefore I_m=\frac{2I_0}{π}=0.637I_0\]

Hence, mean value of ac over positive half cycle is $0.637$ times the peak value of ac i.e. $63.7\%$ of the peak value.

In the same way, the mean value of ac over negative half cycle is obtained by integrating $\text{(2)}$ from $t=\frac{T}{2}$ to $t=0$. By doing so, we get $-0.637I_0$. Hence, over one complete cycle, mean value of ac is $0.637I_0$$-0.637I_0$$=0$.

Similarly, the mean value of alternating emf can be obtained and we get,

\[E_m=2\frac{E_0}{π}=0.637E_0\]

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