在新加坡，公用事业公司向家庭提供频率为 50 Hz（即每秒 50 个周期）的交流（AC）电源。在每个周期中，交流电压以正弦方式在 230 伏和 -230 伏之间振荡。电源电压 $V$ 可以由以下方程表示： $V = 230\sin(100\pi t)$，其中 $t$ 是以秒为单位的时间。

1. Identify the coefficient of $t$ in the sine function and calculate the period

The function is $V = 230\sin(100\pi t)$.
The coefficient of $t$ is $b = 100\pi$.
The period $p = \frac{2\pi}{b} = \frac{2\pi}{100\pi} = \frac{1}{50} = 0.02$.

1. Set up the equation for the maximum voltage

The maximum voltage is $230$ V. We set $V = 230$:
$230\sin(100\pi t) = 230$
$\sin(100\pi t) = 1$

2. Solve for the first time $t$

The sine function first reaches $1$ when the angle is $\frac{\pi}{2}$ radians.
$100\pi t = \frac{\pi}{2}$
$t = \frac{\pi}{2 \times 100\pi}$
$t = \frac{1}{200} = 0.005$

1. Identify the amplitude, period, and key points of the graph

From the equation $V = 230\sin(100\pi t)$, the amplitude is $230$.
From part (i), the period $p = 0.02$ s.
The key points for one cycle of a sine graph are at $t = 0, \frac{p}{4}, \frac{p}{2}, \frac{3p}{4}, p$.
These correspond to $t = 0, 0.005, 0.01, 0.015, 0.02$.
The corresponding $V$ values are $0, 230, 0, -230, 0$.

2. Sketch the graph

Draw a smooth sine curve passing through $(0,0)$, $(0.005, 230)$, $(0.01, 0)$, $(0.015, -230)$, and $(0.02, 0)$. Label the $V$-axis with $230$ and $-230$, and the $t$-axis with the key time values.

1. Set up the equation for the voltage reaching 115 volts

$230\sin(100\pi t) = 115$

2. Simplify the equation to find the basic angle

$\sin(100\pi t) = \frac{115}{230} = \frac{1}{2}$
Basic angle $\alpha = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$

3. Solve for $t$ within the first cycle

For the first cycle, $100\pi t$ is in the first and second quadrants.
$100\pi t = \frac{\pi}{6}, \pi - \frac{\pi}{6}$
$100\pi t = \frac{\pi}{6}, \frac{5\pi}{6}$
$t = \frac{1}{600}, \frac{5}{600}$
$t = \frac{1}{600}, \frac{1}{120}$