Question

2.
The average value of the voltage V for a complete cycle that varies with time t as shown below is:



Kindly answer​

Answer 1

Given:

The voltage V for a complete cycle that varies with time t as shown in the graph.

To find:

Average voltage .

Calculation:

General expression of average voltage for an alternating circuit is given as follows:

$\boxed{\: V_{avg} = \dfrac{ \displaystyle\int V \: dt}{ \displaystyle \int \: dt} }$

Now , as per the graph :

• V = V_(0) when 0 < t < T/2

• V = 0 when T/2 < t < T

So , continuing with the integration:

$= > \: V_{avg} = \dfrac{ \displaystyle\int_{0}^{T}V \: dt}{ \displaystyle \int_{0}^{T} \: dt}$

$= > \: V_{avg} = \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} }(V_{0} \: dt )+ \int_{ \frac{T}{2}}^{T}(0 \: dt)}{ \displaystyle \int_{0}^{T} \: dt}$

$= > \: V_{avg} = \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} }(V_{0} \: dt )+ 0}{ \displaystyle \int_{0}^{T} \: dt}$

$= > \: V_{avg} = \dfrac{ \displaystyle\int_{0}^{ \frac{T}{2} }(V_{0} \: dt )}{ \displaystyle \int_{0}^{T} \: dt}$

$= > \: V_{avg} = \dfrac{ \displaystyle \: V_{0} \: \int_{0}^{ \frac{T}{2} } dt }{ \displaystyle \int_{0}^{T} \: dt}$

$= > \: V_{avg} = \dfrac{ \: V_{0} (\frac{T}{2} - 0)}{ T - 0}$

$= > \: V_{avg} = \dfrac{ \: V_{0} (\frac{T}{2} )}{ T }$

$= > \: V_{avg} = \dfrac{ \: V_{0} }{ 2 }$

So, final answer is:

$\boxed{ \bf{ \large{\: V_{avg} = \dfrac{ \: V_{0} }{ 2 } }}}$