To determine the time needed for a sine wave AC supply to reach a certain voltage, we need to consider the formula for a sine wave:
V(t) = Vmax * sin(2πft + φ)
where: V(t) is the voltage at time t, Vmax is the maximum voltage, f is the frequency of the AC supply, t is the time, and φ is the phase angle.
In this case, the maximum voltage (Vmax) is given as 240 volts, and the frequency (f) is 50 Hz.
Let's assume that the phase angle (φ) is 0 for simplicity. We can then rewrite the equation as:
V(t) = 240 * sin(2π * 50t)
To find the time needed for the voltage to reach 200 volts, we can set V(t) equal to 200 and solve for t:
200 = 240 * sin(2π * 50t)
Divide both sides by 240:
0.8333 = sin(2π * 50t)
Now, we can use inverse sine (arcsine) to find the angle:
arcsin(0.8333) = 2π * 50t
Using a calculator, the arcsine of 0.8333 is approximately 56.28 degrees.
Now, we can convert the angle to time by dividing it by the frequency (50 Hz) and the full period (360 degrees or 2π radians):
56.28 degrees / 360 degrees = t / (1 / 50 Hz * 2π)
0.1563 = t / (1 / (50 Hz * 2π))
Simplifying the equation:
0.1563 = t / (100π Hz)
Rearranging the equation to solve for t:
t = 0.1563 * (100π Hz)
Calculating the value:
t ≈ 49.18 milliseconds
Therefore, it takes approximately 49.18 milliseconds for the sine wave AC supply to reach 200 volts from 0 going positive.