1. **Problem:** An alternating current has a periodic time of 25ms and a maximum value of 20A. When time $t=0$, current $i=-10$ amperes. Express the current $i$ in the form $i = A \sin(\omega t \pm \alpha)$. 2. **Formula and rules:** The general form of a sinusoidal current is $$i = A \sin(\omega t \pm \alpha)$$ where $A$ is the amplitude (maximum current), $\omega$ is the angular frequency, and $\alpha$ is the phase angle. - Angular frequency $\omega = \frac{2\pi}{T}$ where $T$ is the period. - At $t=0$, $i = A \sin(\pm \alpha)$. We use this to find $\alpha$. 3. **Calculate angular frequency:** Given $T = 25$ ms = 0.025 s, $$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.025} = 80\pi \text{ rad/s}$$ 4. **Use initial condition to find phase angle $\alpha$:** At $t=0$, $$i = -10 = 20 \sin(\pm \alpha)$$ $$\sin(\pm \alpha) = \frac{-10}{20} = -0.5$$ 5. **Solve for $\alpha$:** $$\pm \alpha = \sin^{-1}(-0.5) = -\frac{\pi}{6} \text{ or } -30^\circ$$ 6. **Choose sign for phase angle:** We can write $$i = 20 \sin(80\pi t - \frac{\pi}{6})$$ This satisfies the initial condition. **Final answer:** $$\boxed{i = 20 \sin\left(80\pi t - \frac{\pi}{6}\right)}$$