Question

If the peak value of an alternating emf is 10 V what is its mean value over half cycle.

Answer 1

Given:

Peak value of alternating EMF = 10 volts.

To find:

Average value of half cycle?

Calculation:

So, alternating EMF is :

$V = V_{0} \sin( \omega t)$

Now, average value is :

$V_{avg} = \dfrac{ \displaystyle\int_{0}^{T/2} V dt}{\displaystyle \int_{0}^{T/2} dt}$

$\rm \implies V_{avg} = \dfrac{ \displaystyle\int_{0}^{T/2} V dt}{ \dfrac{T}{2} }$

$\rm \implies V_{avg} = \dfrac{ \displaystyle\int_{0}^{T/2} V_{0} \sin( \omega t) dt}{ \dfrac{T}{2} }$

$\rm \implies V_{avg} = \dfrac{2V_{0} \displaystyle\int_{0}^{T/2} \sin( \omega t) dt}{T }$

$\rm \implies V_{avg} = \dfrac{ - 2V_{0} \displaystyle \dfrac{\cos( \omega t)}{ \omega} \bigg|_{0}^{T/2} }{T }$

$\rm \implies V_{avg} = \dfrac{ - 2V_{0} \displaystyle \cos( \omega t) \bigg|_{0}^{T/2} }{ \omega T }$

• Put $\omega$ = 2π/T

$\rm \implies V_{avg} = \dfrac{ 2V_{0} \displaystyle \bigg \{\cos(0) - \cos(\pi) \bigg \} }{ 2\pi}$

$\rm \implies V_{avg} = \dfrac{ 2V_{0} \displaystyle \bigg \{1 + 1 \bigg \} }{ 2\pi}$

$\rm \implies V_{avg} = \dfrac{ 2V_{0} \displaystyle \bigg \{2 \bigg \} }{ 2\pi}$

$\rm \implies V_{avg} = \dfrac{ 2V_{0} }{ \pi}$

$\rm \implies V_{avg} = \dfrac{ 2 \times 10 }{ \pi}$

$\rm \implies V_{avg} = \dfrac{ 20 }{ \pi}$

$\rm \implies V_{avg} = 6.36 \: volt$

So, average EMF over half cycle is 6.36 Volt