A full cycle is $36\text{ ms}$, so the half-cycle (positive or negative) is $18\text{ ms}$.

The sine wave crosses zero at $0\text{ ms}$, $18\text{ ms}$, and $36\text{ ms}$.

To reach $-125\text{ V}$, the time must be greater than the zero-crossing at $18\text{ ms}$.

Solving $\frac{\pi}{18} t = \frac{7\pi}{6}$ gives $t = 21\text{ ms}$.

Predictable behavior of sinusoidal circuits

Standardized analysis for AC system design

Calculation requires knowledge of trigonometric quadrants

Potential for error if the wrong cycle quadrant is selected

Power system waveform analysis

Electronics signaling and timing circuits

The value $-125\text{ V}$ is first reached in the negative half-cycle after the zero crossing at $18\text{ ms}$. Adding $3\text{ ms}$ (the duration from zero to $-V_m/2$ in the positive cycle) to $18\text{ ms}$ yields $21\text{ ms}$.

Option A ($18\text{ ms}$) is a zero-crossing point. Option B ($9\text{ ms}$) is the peak of the positive cycle ($+250\text{ V}$).