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Average Value of Alternating Current | What is the Average Value of Alternating Current Over A Complete Cycle

Average Value of Alternating Current | What is the Average Value of Alternating Current Over A Complete Cycle :- As we have seen in the article on alternating current, the value of alternating current is positive during the first half cycle (t = 0 to t = T/2) and equally negative during the second half cycle (t = T/2 to t = T). Therefore, the average value of alternating current over one complete cycle is zero. However, since the first and second half cycles are symmetrical, we can determine the average value of the alternating current for any one half cycle.

The average (mean) value of an alternating current over any one half-cycle is equal to that steady current which would transfer the same amount of charge through an electric circuit in half a cycle (time T/2) as the alternating current transfers through the same circuit in that half-cycle.

To find the average value of alternating current, let the alternating current flowing in an electrical circuit be given by the following function :-

$\displaystyle I={{I}_{0}}sin\text{ }\omega t$ …..(1)

Where I₀ is the peak value of the alternating current.

If the value of alternating current is considered constant for a small time interval dt, then the small amount of charge dq flowing in the circuit is :

$\displaystyle dq=Idt=\left( {{I}_{0}}sin\text{ }\omega t \right)dt$ …..(2)

Total charge flowing through the circuit during the first half cycle (t = 0 से t = T/2) :

$\displaystyle q=\int\limits_{0}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}{\left( {{I}_{0}}sin\text{ }\omega t \right)dt}$

$\displaystyle \Rightarrow q={{I}_{0}}\int\limits_{0}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}{\left( sin\text{ }\omega t \right)dt}$

$\displaystyle \Rightarrow q={{I}_{0}}\left[ \frac{-cos\text{ }\omega t}{\omega } \right]_{0}^{{}^{T}\!\!\diagup\!\!{}_{2}\;}$

$\displaystyle \Rightarrow q=-\frac{{{I}_{0}}}{\omega }\left[ cos\text{ }\left( \omega \times \frac{T}{2} \right)-cos\text{ }\left( \omega \times 0 \right) \right]$

$\displaystyle \Rightarrow q=-\frac{{{I}_{0}}}{\omega }\left[ cos\text{ }\left( \frac{2\pi }{T}\times \frac{T}{2} \right)-cos\text{ }\left( 0 \right) \right]$

$\displaystyle \Rightarrow q=-\frac{{{I}_{0}}}{\omega }\left[ cos\text{ }\left( \pi \right)-cos\text{ }\left( 0 \right) \right]$

$\displaystyle \Rightarrow q=-\frac{{{I}_{0}}}{\omega }\left[ -1-1 \right]$

$\displaystyle \Rightarrow q=\frac{2{{I}_{0}}}{\omega }$ …..(3)

If I_m is the average value of the alternating current in the first half cycle (from t = 0 to t = T/2), then

$\displaystyle q={{I}_{m}}\times \frac{T}{2}$ …..(4)

From equations (3) and (4),

$\displaystyle {{I}_{m}}\times \frac{T}{2}=\frac{2{{I}_{0}}}{\omega }$

$\displaystyle \Rightarrow {{I}_{m}}\times \frac{T}{2}=\frac{2{{I}_{0}}}{{}^{2\pi }\!\!\diagup\!\!{}_{T}\;}$

$\displaystyle \therefore {{I}_{m}}=\frac{2}{\pi }{{I}_{0}}=\text{0}\text{.637}{{I}_{0}}$ …..(5)

Thus, in a half cycle, the average value of alternating current is 0.637 times its peak value (I₀), or 63.7% of the peak value.

Similarly, if we integrate equation (2) for the second half cycle (from t = T/2 to t = T), then we obtain the following value :

$\displaystyle {{I}_{m}}=-\frac{2}{\pi }{{I}_{0}}=-\text{0}\text{.637}{{I}_{0}}$ …..(6)

Thus, over one complete cycle (from t = 0 to t = T), the average value of alternating current becomes zero (0.637 I₀ – 0.637 I₀ = 0).

The above result can also be obtained by integrating equation (2) within the limits from t = 0 to t = T.

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Average Value of Alternating Current | What is the Average Value of Alternating Current Over A Complete Cycle

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