An AC source is rated 222V and 60Hz. The average voltage is calculated in a time interval of 16.66ms is?
a) Must be zero
b) may be zero
c) is never zero
d) is $11\sqrt{2}V$

Hint: The average value of voltage in the complete cycle of AC is zero. It is given in the question that the frequency of the AC source is 60Hz. Hence we will first determine the time period of the AC source i.e. equal to the reciprocal of the frequency. Further we will compare the time period of the frequency of the AC source with the given time interval and accordingly determine what should be the value of the average voltage of the AC.

Formula used:
$f=\dfrac{1}{T}$
$V={{V}_{\circ }}\operatorname{Sin}\left( 2\pi ft \right)$

Complete step-by-step answer:
In the above question it is given to us that the frequency of the AC source is f= 60Hz. Hence the time interval (T) of the Ac source is,
$\begin{align}
  & f=\dfrac{1}{T} \\
 & T=\dfrac{1}{f}\text{, since f=60Hz} \\
 & \text{T=}\dfrac{1}{60}s=0.01666=16.6ms \\
\end{align}$
In the question it is asked to calculate the average voltage during this time interval. This time interval corresponds to the time taken by the Ac source to complete it’s one complete cycle. The average value of AC along one complete cycle is 0. Therefore we can conclude that the average voltage during the time interval of 16.66ms will be zero.

So, the correct answer is “Option a)”.

Note: In general the is we want to calculate the instantaneous voltage at any instant t, is given by $V={{V}_{\circ }}\operatorname{Sin}\left( 2\pi ft \right)$ where ${{V}_{\circ }}$ is the maximum value of the Ac voltage in a complete cycle, f is the frequency and t is the instant of time for which the instantaneous voltage is to be determined. Therefore we can divide the time interval of the instantaneous voltage and obtain the average voltage.