Question

An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It

(a) must be zero
(b) may be zero
(c) is never zero
(d) $\mathrm{is}\left(220/\sqrt{2}\right)\mathrm{V}$

Answer 1

(b) may be zero

Let the AC voltage be given by,
$V=V_0\sin \omega t$
Here, ω = 2$\pi$f = 314 rad/s
The average voltage over the given time,
$$\begin{gathered} V_{avg}=\frac{{\int }_0^{0.01}Vdt}{{\int }_0^{0.01}dt}=-V_0{\left[\frac{\cos \omega t}{\omega }\right]}_0^{0.01} \\ =\frac{V_0}{\omega \times 0.01}\left(1-\cos \omega \left(0.01\right)\right) \\ =\frac{V_0}{314\times 0.01}\left(1-\cos \left(314\times 0.01\right)\right) \\ =\frac{V_0}{3.14}\left(1-\mathrm{cos\pi }\right) \\ =\frac{2V_0}{\mathrm{\pi }}=140.127\mathrm{V} \end{gathered}$$

Also, when $V=V_0\cos \omega t$,
$$\begin{gathered} V_{avg}=\frac{{\int }_0^{0.01}Vdt}{{\int }_0^{0.01}dt}=V_0{\left[\frac{\sin \omega t}{\omega }\right]}_0^{0.01} \\ =\frac{V_0}{\omega \times 0.01}\left(\sin \omega \left(0.01\right)-0\right) \\ =\frac{V_0}{314\times 0.01}\left(\sin \left(314\times 0.01\right)\right) \\ =\frac{V_0}{3.14}\left(\sin \pi \right) \\ =0 \end{gathered}$$

From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.