An alternating voltage is represented as \(\mathrm{E}=20 \sin 300 \mathrm{t}\). The average value of voltage over one cycle will be. (a) zero (b) 10 volt (c) \(20 \sqrt{2}\) volt (d) \((20 / \sqrt{2})\) volt

In the given voltage function, E(t) = 20 sin(300t), the angular frequency ω is 300 rad/s. To find the time period (T) of the alternating voltage, we can use the formula: T = \(\frac{2\pi}{ω}\) Now, plug in the value of ω: T = \(\frac{2\pi}{300}\) s

To find the average value of the voltage over one cycle, we will integrate the voltage function E(t) over one time period and then divide the result by the time period. Since the voltage is alternating, we need to find the average of the absolute value of the voltage function: Average Voltage = \(\frac{1}{T} \int_0^T |E(t)| dt\) First, substitute the given voltage function and the calculated time period T: Average Voltage = \(\frac{1}{\frac{2\pi}{300}} \int_0^{\frac{2\pi}{300}} |20 \sin(300t)| dt\) Now, the absolute value of sin function varies from 0 to 1, so the absolute value of the given voltage function is simply the voltage function itself: Average Voltage = \(\frac{1}{\frac{2\pi}{300}} \int_0^{\frac{2\pi}{300}} 20 \sin(300t) dt\) Integrate the function: Average Voltage = \(\frac{1}{\frac{2\pi}{300}} [-\frac{1}{15}\cos(300t)]_0^{\frac{2\pi}{300}}\) Evaluate the definite integral: Average Voltage = \(\frac{1}{\frac{2\pi}{300}} [-\frac{1}{15}\cos(300(\frac{2\pi}{300})) + \frac{1}{15}\cos(0)]\) Average Voltage = \(\frac{300}{2\pi} [\frac{1}{15}(1-1)]\) Average Voltage = 0 So the correct answer is (a) zero.