Question

An alternating voltage, v, has a periodic time of 50 ms and a peak value of 80 V. When the time t = 0, v = −20 V

Express the instantaneous voltage v in the form

v = Asin(ωt ± θ)

Answer 1

The AC voltage can be expressed in the form v = Asin(ωt ± θ), where A is the amplitude of the voltage, ω is the angular frequency, t is the time, and θ is the phase angle. To express the given voltage in this form, we determine the values of A, ω, and θ. The instantaneous voltage can be expressed as v = 80*sin(40πt - 14.48°).

Step-by-step explanation:

The AC voltage can be expressed in the form v = Asin(ωt ± θ), where A is the amplitude of the voltage, ω is the angular frequency, t is the time, and θ is the phase angle. To express the given voltage in this form, we need to determine the values of A, ω, and θ.

The peak value of the voltage, 80 V, is equal to A, so A = 80 V. The periodic time, 50 ms, is the time for one complete cycle of the voltage waveform, which is equal to the period 2π/ω. Therefore, 50 ms = 2π/ω. Solving for ω, we find ω = 2π/50 ms = 40π rad/s.

Since the voltage is given as v = -20 V at t = 0, θ can be found by substituting the given values into the voltage equation: -20 V = 80 V*sin(ω*0 + θ). This simplifies to -1/4 = sin(θ). Taking the inverse sin of both sides, we find θ = sin-1(-1/4) = -14.48°.

Therefore, the instantaneous voltage can be expressed as v = 80*sin(40πt - 14.48°).