Question

An AC source is rated 222 V, 60 Hz. The average voltage is calculated in a time interval of 16.67 ms.

• A. Must be zero
• B. May be zero
• C. Is never zero
• D. Is $(111\sqrt{2})$ V

Answer 1

Answer: (B)

May be zero

Answer 2

The period of the AC source is $T = \frac{1}{f} = \frac{1}{60}$ s, which is approximately $16.666...$ ms. The given time interval is $\Delta t = 16.67$ ms. Since $\Delta t$ is not exactly an integer multiple of the period $T$, the average voltage over this interval is not necessarily zero. However, the instantaneous voltage is given by $v(t) = V_0 \sin(\omega t + \phi)$. The average voltage over the interval $[t_1, t_1 + \Delta t]$ is $V_{avg} = \frac{1}{\Delta t} \int_{t_1}^{t_1+\Delta t} v(t) dt$. This integral evaluates to $V_{avg} = \frac{V_0}{\omega \Delta t} \left[ \cos(\omega t_1 + \phi) - \cos(\omega (t_1+\Delta t) + \phi) \right]$. While $\omega \Delta t \approx 2.0004\pi$, which is not an exact multiple of $2\pi$, it is possible to choose the initial phase $\phi$ and starting time $t_1$ such that $V_{avg} = 0$. Therefore, the average voltage may be zero.