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Mean or Average Value of Alternating emf

Mean or Average Value of Alternating emf :- Like alternating current, the value of alternating electromotive force (emf) 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, over one complete cycle, the average value of alternating emf is zero. Hence, we will determine the average value of the alternating emf for any one half cycle.

The mean or average value of alternating emf over any half cycle is equal to that value of a direct emf (constant potential difference) which would transfer the same amount of charge through an electric circuit in a half cycle (time T/2) as is transferred by the alternating emf through the same circuit in that half cycle.

To find the average value of alternating emf, suppose that the alternating emf applied to an electric circuit is given by the following function :-

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

Where E₀ is the peak (or maximum) value of the alternating emf.

If at any time t the value of current in the circuit is I, then

$\displaystyle I=\frac{E}{R}=\frac{{{E}_{0}}}{R}sin\text{ }\omega t$

Where R is the resistance of the circuit.

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

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

Total charge flowing in the circuit during the first half cycle (from t = 0 to t = T/2) :

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

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

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

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

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

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

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

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

If E_m is the average value of the alternating emf during the first half-cycle (from t = 0 to t = T/2), then

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

From equations (3) and (4),

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

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

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

Thus, in a half cycle, the mean or average value of alternating emf is 0.637 times its peak value (E₀), 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 :

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

Thus, over one complete cycle (from t = 0 to t = T), the average value of alternating emf 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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Mean or Average Value of Alternating emf

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