The output voltage of AC source is V(t) = (25V)sin(5πt). what is the rms value of the voltage? answer: 18V, but how???
recall that peak voltage (amplitude) of a sine wave is √2 times rms.
25/√2 = 17.67 ≈ 18
To determine the root mean square (rms) value of the given voltage waveform, you need to follow the following steps:
Step 1: Find the square of the voltage function.
The voltage function is given by V(t) = (25V)sin(5πt).
By squaring the voltage function, you get V^2(t) = (25V)^2sin^2(5πt).
Step 2: Find the average of the squared voltage over one complete cycle.
Since the voltage function is a periodic waveform, you need to find the average value of the squared voltage over one complete cycle. In this case, the period T is the time it takes for the waveform to complete one full cycle. For a sine function, the period is 2π divided by the angular frequency (ω).
The angular frequency ω = 5π.
Therefore, T = 2π/ω = 2π/(5π) = 2/5 seconds.
Now, integrate the squared voltage function over one complete cycle:
Integrate V^2(t) from t = 0 to t = T.
∫[0 to T] (25V)^2sin^2(5πt) dt
Since sin^2(5πt) has an average value of 1/2 over one complete cycle, the integral simplifies to:
(25V)^2 * (1/2) * T
= (25V)^2 * (1/2) * (2/5)
= (25V)^2 * (1/5)
= (625V^2) * (1/5)
= 625V^2 / 5
Step 3: Take the square root of the average squared voltage.
The rms value of the voltage (Vrms) is the square root of the average of the squared voltage:
Vrms = √(625V^2 / 5)
= √(125V^2)
= √(25V) * √(5V)
= 5V * √(5V)
= 5√5V ≈ 11.18V
Therefore, the rms value of the given voltage waveform is approximately 11.18V, not 18V.