Final answer:
To plot the instantaneous power for one period of the voltage, one must multiply the voltage and current sinusoidal functions, taking into account the phase difference, and visualize the resulting function over one period.
Step-by-step explanation:
The student has provided the equations for voltage and current in an AC circuit with sinusoidal functions. To plot the instantaneous power for one period of the voltage, we can use the formula p(t) = i(t)v(t), where p(t) is the instantaneous power, i(t) is the current, and v(t) is the voltage. Given the functions v(t) = 15.4sin(754t - 30°) and i(t) = 6.35sin(754t + 15°), we need to multiply these two equations to find p(t), the instantaneous power.
To compute p(t), we multiply the magnitude of the voltage by the magnitude of the current and apply the trigonometric identity for the product of two sinusoidal functions while taking into account their phase difference. This results in a new sinusoidal function that represents the varying power over time. The phasor diagram would visualize the phase difference between current and voltage.
The final step is to plot this new function over one period to observe how the power changes over time, keeping in mind that the period is T = 2π/ω, where ω is the angular frequency. In our case, ω = 754 rad/s, and thus the period T can be calculated. The resulting plot would show periods where the power is positive, indicating power is being dissipated, and periods where the power may be negative, indicating power is being returned to the source or stored.