Question

An AC voltage is given by $$E=E_0\sin {\displaystyle \frac{2\pi t}{T}}$$. Then the mean voltage is calculated over the interval of $${\displaystyle \frac{T}{2}}$$ second.
A. is always zero
B. is never zero
C. is always $${\displaystyle \frac{E_0}{2\pi }}$$
D. maybe zero

Answer 1

Hint: Use the formula for the mean voltage over the time interval. This formula gives the integration of voltage over the time interval giving the relation between the voltage and time interval. Integrate the equation with respect to time correctly to get the final value of the mean voltage.

Formula used:
The formula for the mean voltage over the time interval $$T$$ is
$$\text{Mean voltage}={\displaystyle \frac{\underset{0}{\overset{T}{\int }}Edt}{\underset{0}{\overset{T}{\int }}dt}}$$ …… (1)
Here, $$E$$ is the voltage and $$T$$ is the time interval.

Complete step by step answer:
We have given that the AC voltage is given by
$$E=E_0\sin {\displaystyle \frac{2\pi t}{T}}$$
This AC voltage changes with time.
We have to determine the mean voltage over the time interval of $${\displaystyle \frac{T}{2}}$$ second.
Substitute $$E_0\sin {\displaystyle \frac{2\pi t}{T}}$$ for $$E$$ and $${\displaystyle \frac{T}{2}}$$ for $$T$$ in equation (1).
$$\text{Mean voltage}={\displaystyle \frac{\underset{0}{\overset{{\displaystyle \frac{T}{2}}}{\int }}\left[E_0\sin {\displaystyle \frac{2\pi t}{T}}\right]dt}{\underset{0}{\overset{{\displaystyle \frac{T}{2}}}{\int }}dt}}$$
$$\Rightarrow \text{Mean voltage}={\displaystyle \frac{E_0{\left[{\displaystyle \frac{-\cos {\displaystyle \frac{2\pi t}{T}}}{{\displaystyle \frac{2\pi }{T}}}}\right]}_0^{{\displaystyle \frac{T}{2}}}}{{\left[t\right]}_0^{{\displaystyle \frac{T}{2}}}}}$$
$$\Rightarrow \text{Mean voltage}={\displaystyle \frac{E_0{\displaystyle \frac{T}{2\pi }}\left[-\cos {\displaystyle \frac{2\pi {\displaystyle \frac{T}{2}}}{T}}+\cos {\displaystyle \frac{2\pi 0}{T}}\right]}{\left[{\displaystyle \frac{T}{2}}-0\right]}}$$
$$\Rightarrow \text{Mean voltage}={\displaystyle \frac{E_0{\displaystyle \frac{T}{2\pi }}\left[-\cos \pi +\cos 0\right]}{{\displaystyle \frac{T}{2}}}}$$
$$\Rightarrow \text{Mean voltage}={\displaystyle \frac{E_0}{\pi }}\left[-1+1\right]$$
$$\Rightarrow \text{Mean voltage}=0$$
If the value of the voltage is something other, this value of the mean voltage may not be zero.
Therefore, the mean voltage calculated over the interval of $${\displaystyle \frac{T}{2}}$$ second may be zero.

So, the correct answer is “Option D”.

Additional Information:
The mean voltage over a time interval can also be determined by multiplying the peak voltage by the constant 0.637.
The mean voltage can also be determined by the graphical method with good accuracy.

Note:
The students should keep in mind that the integration of sine of any angle is equal to the negative of the cosine of that angle. Also when the limits of integration are substituted there is a negative sign between the two values. The students should not get confused between this negative sign and the negative sign of the cosine of angle.