Question

The average value of current given by $ {\text{i = }}{{\text{I}}_{\text{m}}}{\text{ sin }}\omega {\text{t}} $, from $ {\text{t = }}\dfrac{\pi }{{2\omega }} $ to $ {\text{t = }}\dfrac{{3\pi }}{{2\omega }} $ is how many times of $ {{\text{I}}_{\text{m}}} $?

Answer 1

Hint: To find the number of times the average value becomes with respect to the given time interval, we calculate the integral of the general form of an average of a function in the interval and compare it to $ {{\text{I}}_{\text{m}}} $.

Complete step by step answer:
We integrate the sine function and substitute the intervals given in the above. The integration of a sine function is given by, sin x = - cos x.
Given data,
$ {\text{i = }}{{\text{I}}_{\text{m}}}{\text{ sin }}\omega {\text{t}} $
Time interval $ {\text{t = }}\dfrac{\pi }{{2\omega }} $ to $ {\text{t = }}\dfrac{{3\pi }}{{2\omega }} $ .
The average value of a function ‘f’ in a certain time interval from ‘a’ to ‘b’ is given by the formula –
$ {{\text{f}}_{{\text{avg}}}} = \dfrac{{\int\limits_{\text{a}}^{\text{b}} {{\text{fdt}}} }}{{\int\limits_{\text{a}}^{\text{b}} {{\text{dt}}} }} $ .
Here we are supposed to find the average value of current ‘i’ from a time interval $ {\text{t = }}\dfrac{\pi }{{2\omega }} $ to $ {\text{t = }}\dfrac{{3\pi }}{{2\omega }} $

 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = }}\dfrac{{\int\limits_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}} {{\text{idt}}} }}{{\int\limits_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}} {{\text{dt}}} }}{\text{ }} $
Given value of current is $ {\text{i = }}{{\text{I}}_{\text{m}}}{\text{ sin }}\omega {\text{t}} $ , we substitute this in the above equation we get
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = }}\dfrac{{\int\limits_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}} {\left( {{{\text{I}}_{\text{m}}}{\text{ sin }}\omega {\text{t}}} \right){\text{dt}}} }}{{\int\limits_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}} {{\text{dt}}} }}{\text{ }} $
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = }}\dfrac{{{{\text{I}}_{\text{m}}} \times \dfrac{{\left[ { - {\text{cos}}\omega {\text{t}}} \right]_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}}}}{\omega }}}{{\left[ {\text{t}} \right]_{\dfrac{\pi }{{2\omega }}}^{\dfrac{{3\pi }}{{2\omega }}}}}{\text{ }} $
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = }}\dfrac{{{{\text{I}}_{\text{m}}} \times \dfrac{{\left( { - 1} \right) \times \left[ {{\text{cos}}\omega \times \dfrac{{3\pi }}{{2\omega }} - {\text{cos}}\omega \times \dfrac{\pi }{{2\omega }}} \right]}}{\omega }}}{{\left[ {\dfrac{{3\pi }}{{2\omega }} - \dfrac{\pi }{{2\omega }}} \right]}}{\text{ }} $
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = - }}\dfrac{{{{\text{I}}_{\text{m}}}}}{\pi } \times \left[ {{\text{cos}}\dfrac{{3\pi }}{2} - {\text{cos}}\dfrac{\pi }{2}} \right]{\text{ }} $
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = - }}\dfrac{{{{\text{I}}_{\text{m}}}}}{\pi } \times \left[ {0 - 0} \right]{\text{ }} $
 $ \Rightarrow {{\text{i}}_{{\text{avg}}}}{\text{ = 0}} $
Hence the average value of current is zero times of $ {{\text{I}}_{\text{m}}} $ .

Note:
In order to answer this type of question the key is to know the general formula of average value of a function using integration.
Integral of a function of the form $ \int\limits_{\text{a}}^{\text{b}} {{\text{dx}}} $ is given by $ \left[ {\text{x}} \right]_{\text{a}}^{\text{b}} = \left[ {{\text{b - a}}} \right] $ .
Integral of a function of the form $ \int\limits_{\text{p}}^{\text{q}} {{\text{sin}}\left( {{\text{bx}}} \right)} {\text{ dx}} $ is given by $ \left[ {\dfrac{{{\text{ - cosbx}}}}{{\text{b}}}} \right]_{\text{p}}^{\text{q}} $ .
Also the values of $ {\text{cos}}\dfrac{{3\pi }}{2}{\text{ and cos}}\dfrac{\pi }{2} $ is equal to zero and is obtained from the trigonometric table of cosine function.