Question

The variation in an alternating emf with time is shown below. Calculate the mean value per cycle of the given emf.

• A. $\frac{V_0}{2}$
• B. $0$
• C. $\frac{V_0}{\sqrt{2}}$
• D. $\frac{V_0}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{V_0}{2}$

From the given graph, equation for fluctuating emf is given as

$V=\{\begin{array}{ccc}& \frac{2V_0}{T}t;& 0<t\le T/2\\ & & \\ & \frac{-2V_0}{T}t+2V_0;& T/2<t\le T\end{array}$

So,
$V_{mean}=\frac{1}{T}\left\{{\int }_0^{T/2}\frac{2V_0}{T}tdt+{\int }_{T/2}^T(2V_0-\frac{2V_0}{T}t)dt\right\}$

$=\frac{1}{T}\{\frac{V_0}{T}\left[\frac{T^2}{4}\right]+2V_0\frac{T}{2}-\frac{V_0}{T}[T^2-\frac{T^2}{4}]\}$

$=\frac{1}{T}\{\frac{V_{0}T}{4}+V_{0}T-\frac{3}{4}{V}_{0}T\}$

$V_{mean}=\frac{1}{T}\{\frac{5}{4}{V}_{0}T-\frac{3}{4}{V}_{0}T\}$

$\therefore V_{mean}=\left(\frac{V_0}{2}\right)$

Hence, option $\left(\text{A}\right)$ is correct.

Alternative Method $V_{mean}=\frac{{\int }_0^{T}Vdt}{{\int }_0^{T}dt}$ Here, ${\int }_0^{T}Vdt$ is the area under the curve. So, $V_{mean}=\frac{\text{Area in one time period}}{T}$ $=\frac{1/2\times T\times V_0}{T}=\frac{V_0}{2}$