Q.An alternating voltage is given by e = 8 sin 628.4t. Find

Compare with standard form and evaluate at the given instant.

Given $e = 8\sin(628.4\,t)$, matched to $e=e_0\sin(\omega t)$: $e_0=8$ V, $\omega=628.4$ rad/s.

1. Peak value: $e_0 = 8$ V.
2. Frequency: $f = \dfrac{\omega}{2\pi} = \dfrac{628.4}{6.2832} \approx 100$ Hz. (iii) Instantaneous value at $t=10$ ms $=0.01$ s: $\omega t = 628.4\times0.01 = 6.284\ \text{rad}$ Since $2\pi \approx 6.2832$ rad, this is just past one complete cycle ($t\approx T=10$ ms is almost exactly the period), so $\omega t - 2\pi \approx 0.0008$ rad: …