Question

If the A.C. main supply is of $220V$then the average e.m.f. during positive half cycle will be:
(A) $198.2V$
(B) $98.2V$
(C) $9.82V$
(D) Zero

Answer 1

Hint
In the A.C. supply, the value of current is always alternating, it is the waveform of a sine or cos wave, also, after half cycle the current changes its polarity. So the average e.m.f. in a full cycle is zero, but other than that there is presence of a net e.m.f. in the supply. To give a meaning and for calculation, the RMS (root mean square) values of the e.m.f. is used. Using this, we can also figure out the average e.m.f. produced by the supply.
$\Rightarrow {E_{avg}} = \dfrac{{2\sqrt 2 {E_{RMS}}}}{\pi }$

Complete step by step answer
A normal representation of voltage in A.C. supply is given by putting RMS values of the alternating component. All the values displayed or mentioned about AC signals are represented in its RMS form, in this question too, the RMS value of the AC Supply is given as $220V$.
Relating to this, the average value of the supply voltage is given by-
$\Rightarrow {E_{avg}} = \dfrac{{2\sqrt 2 {E_{RMS}}}}{\pi }$
Where ${E_{avg}}$refers to the average e.m.f.
While ${E_{RMS}}$refers to the Root Mean Square of e.m.f.
Putting the value ${E_{RMS}} = 220$we get,
$\Rightarrow {E_{avg}} = \dfrac{{2\sqrt 2 \times 220}}{\pi }$
$\Rightarrow {E_{avg}} = \dfrac{{622.25}}{\pi }$
$\Rightarrow {E_{avg}} = 198.06V$
Hence, option (A) is correct.

Additional Information
The three ways to represent the values of e.m.f. in an Alternating current system are-
-Maximum e.m.f: (${E_0}$) This tells the maximum e.m.f. or the peak amplitude of the sinusoidal curve. At any given instant of time the e.m.f. can be given using-
$\Rightarrow E = {E_0}\operatorname{Sin} (\omega t - \phi )$
Here $\omega $ , $t$ and $\phi $ represent the phase of the supply.
-Average e.m.f (${E_{avg}}$) It is defined as the sum of all the values of e.m.f produced in a cycle or given part of the cycle, divided by the time taken to complete the cycle. It is given by-
$\Rightarrow {E_{avg}} = \dfrac{1}{T}\int\limits_0^T {E(t)dt} $

Where T is the total time taken to complete the cycle.
-RMS e.m.f: (${E_{RMS}}$) To calculate the RMS value, small components of the AC e.m.f are squared and added, after this the mean of these values is taken. This means is then put into the root, to make it practically close to the observed values. This removes the problems caused by negative values in calculations. It is given by-
$\Rightarrow {E_{RMS}} = \sqrt {\dfrac{1}{T}\int\limits_0^T {{E^2}(t)dt} } $

Note
The relation between the 3 representations of e.m.f. in an AC signal is given by-
$\Rightarrow {E_{avg}} = \dfrac{{2{E_0}}}{\pi } = \dfrac{{2\sqrt 2 {E_{RMS}}}}{\pi }$
In everyday use, AC voltages are always given as RMS values because this allows a sensible comparison to be made with steady DC voltages.