Question

Define average value of a.c. over half a cycle. Establish the relationship between the 'average value' and the 'peak value of an alternating current.

Answer 1

Hint : The average value can be determined by adding together a series of instantaneous values of the alternation (between 0° and 180°), and then dividing the sum by the number of instantaneous values used. The charge sent by the alternating current $I$ in time $dt$ is given by,
$dq = {I_0}\sin \omega tdt$ .

Complete step by step answer
Alternating current is positive during the first half cycle and negative during another half cycle, so the mean average value of a.c. over one cycle is zero. We can find the mean or average value of a.c. over any half cycle. Mean or average value of alternating current is that value of steady current which sends the same amount of charge through a circuit in a certain time interval as is sent by an alternating current through the same circuit in the same time interval.
$I = {I_0}\sin \omega t$ where ${I_0}$ is the peak current.
The average value of an alternating current is the average of all the instantaneous values during one alternation. Since the current increases from zero to peak value and decreases back to zero during one alternation, the average value must be some value between those two limits.The average value can be determined by adding together a series of instantaneous values of the alternation (between 0° and 180°), and then dividing the sum by the number of instantaneous values used. The computation would show that one alternation of a sine wave has an average value equal to 0.636 times the peak value.
The charge sent by the alternating current I in time $dt$ is given by,
$dq = Idt$
$\Rightarrow dq = {I_0}\sin \omega tdt$
We should note that, ${I_0}$ is the peak current.
The amount of charge passing through the circuit in half time period can be obtained by integrating above equation (i) from $t = 0$ to $t = {T \mathord{\left/ {\vphantom {T 2}} \right.} 2}$ .
$q = \int\limits_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}} {{I_0}\sin \omega tdt = } {I_0}\int\limits_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}} {\sin \omega tdt}$
$= {I_0}\left[ {\dfrac{{ - \cos \omega tdt}}{\omega }} \right]_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}} = - \dfrac{{{I_0}}}{\omega }\left[ {\cos \omega t} \right]_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}}$
Applying the limits,
- \dfrac{{{I_0}}}{{{\raise0.7ex\hbox{ {2\pi } } \\!\mathord{\left/ {\vphantom {{2\pi } T}}\right.} \\!\lower0.7ex\hbox{ T }}}}\left[ {\cos \dfrac{{2\pi }}{T}t} \right]_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}} = - \dfrac{{{I_0}T}}{{2\pi }}\left[ {\cos \dfrac{{2\pi }}{T}\dfrac{T}{2} - \cos 0} \right]
Simplifying the equation further,
$= - \dfrac{{{I_0}T}}{{2\pi }}\left[ {\cos \pi - \cos 0} \right]$
We know, $\cos \pi = - 1$ and $\cos 0 = 1$ ,
$= - \dfrac{{{I_0}T}}{{2\pi }}\left[ { - 1 - 1} \right] = - \dfrac{{{I_0}T}}{{2\pi }} \times - 2$ .
Thus, $q = \dfrac{{{I_0}T}}{\pi }$
If ${I_m}$ be the mean value of an a.c. over positive half cycle, then the charge sent by it in time T/2 is given by,
$q = {I_m}\dfrac{T}{2} = \dfrac{{{I_0}T}}{\pi }$
The relationship between the 'average value' and the 'peak value of an alternating current is,
$\therefore {I_m} = \dfrac{{2{I_0}}}{\pi } = 0.637{I_0}$ .

Note
Similarly, the mean or average value of a.c. over the negative half cycle is obtained by integrating equation $q = \int\limits_0^{{T \mathord{\left/ {\vphantom {T 2}} \right.} 2}} {{I_0}\sin \omega tdt}$ from $t = {T \mathord{\left/ {\vphantom {T 2}} \right.} 2}$ to $t = T$ . It comes out to be $- 0.637{I_0}$ .
Hence the mean or average value of a.c. over one complete cycle is zero.