Alternating Current, Simple AC Circuits (JEE LEVEL)

Mean Value of AC

Mean value of a time varying current can be found using the following equation

          \displaystyle {{I}_{{av}}}=\frac{{\int{{Idt}}}}{{\int{{dt}}}} the limits of integrations depends upon the time interval in which mean value is required.

Usually the mean value of ac is calculated for two time intervals

(1)     Over half time period (half cycle).

(2)     Over full time period (full cycle).

The average value of ac over half cycle can be calculated as follows:

                   \displaystyle {{I}_{{av}}}              = \displaystyle \frac{{\int\limits_{0}^{{T/2}}{{Idt}}}}{{T/2}}=\frac{{2\int\limits_{0}^{{T/2}}{{I\,dt}}}}{T}                            … (1)

                            = \displaystyle \frac{{2\int\limits_{0}^{{T/2}}{{{{I}_{0}}\sin \omega t\,(dt)}}}}{T}

                            = \displaystyle \frac{{2{{I}_{0}}}}{T}\left[ {-\frac{{\cos \omega t}}{\omega }} \right]_{0}^{{T/2}} = \displaystyle \frac{{2{{I}_{0}}}}{{\omega T}}\left[ {-\cos \omega \frac{T}{2}-(-\cos 0)} \right]

                            = \displaystyle \frac{{2{{I}_{0}}}}{{\frac{{2\pi }}{T}.T}}\left[ {-\cos \frac{{2\pi }}{T}.\frac{T}{2}+\cos 0} \right]

                            = \displaystyle \frac{{{{I}_{0}}}}{\pi }\left[ {-\cos \left( {\frac{{2\pi .T}}{{T.2}}} \right)-(-1)} \right]

                            = \displaystyle \frac{{{{I}_{0}}}}{\pi }\,\,[-\cos \pi +1]

(ii)     Similarly for calculating mean value of alternating current over full cycle we have to replace the limit of integration in equation (1) by 0 to T. Solving integration in the same way we find that the mean value of ac over full cycle is zero. Also looking at graph, the positive current pattern is equal and opposite to negative pattern, hence mean value of ac over full cycle is zero.

RMS value

The rms value of any quantity is “square root of the arithmetic mean of squares of all those values whose rms value is required”. RMS value is also called effective values.

          Thus rms value of 1, 2, 3, 4 is

                   \displaystyle \sqrt{{({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}})/4}} \displaystyle =\sqrt{{\frac{{15}}{2}}}

RMS value of AC

In case of AC, the rms value of an AC comes out to be that value of DC which will produce the same heating effect on passing through the same resistance over the same period of time. Hence rms value of ac is also called Equivalent DC ( \displaystyle {{I}_{{eq}}})

If the magnitude (I) of AC at any instant of time is given by

                   \displaystyle I\text{ }=\text{ }{{I}_{0}}sin\omega t,

          Then heat (H) produced by this AC when passed through a resistance (R) during one time period (T) will given by

                   \displaystyle H=\int\limits_{0}^{T}{{dH}}=\int\limits_{0}^{T}{{{{I}^{2}}\,Rdt}}                                              … (i)

          And during this one time period T, the heat produced by a DC ( \displaystyle {{I}_{{rms}}} or \displaystyle {{I}_{{eq}}}) will be

                                 H  \displaystyle =I_{{rms}}^{2}RT                                                               … (ii)

          Hence, equation (i) and (ii), when equated will give us the value of the equivalent DC or rms value of ac ( \displaystyle {{I}_{{eq}}}).

                      \displaystyle I_{{rms}}^{2}RT  = \displaystyle \int\limits_{0}^{T}{{{{I}^{2}}Rdt}}=\int\limits_{0}^{T}{{{{{({{I}_{0}}\sin \omega t)}}^{2}}R\,\,dt}}

                                      = \displaystyle I_{0}^{2}R\int\limits_{0}^{T}{{{{{\sin }}^{2}}\omega t\,\,dt}}                 

                                      = \displaystyle I_{0}^{2}R\,\,\int\limits_{0}^{T}{{\left( {\frac{{1-\cos 2\omega t}}{2}} \right)}}\,\,dt                     [because \displaystyle \text{cos(2 }\!\!\omega\!\!\text{ t)= 1 — 2 si}{{\text{n}}^{\text{2}}}\text{ }\!\!\omega\!\!\text{ t})

                                      = \displaystyle \frac{{I_{0}^{2}R}}{2}\left[ {\int\limits_{0}^{T}{{dt}}-\int\limits_{0}^{T}{{\cos \,2\omega t\,\,dt}}} \right]

                                      = \displaystyle \frac{{I_{0}^{2}R}}{2}\left[ {T-\left\{ {\frac{{\sin 2\omega t}}{{2\omega }}} \right\}_{0}^{T}} \right]

                                      \displaystyle =\frac{{I_{0}^{2}RT}}{2}

[Because when \displaystyle \text{t = T, sin2 }\!\!\omega\!\!\text{ t = sin 2 }\!\!\omega\!\!\text{ T = sin(2}\text{. (2 }\!\!\pi\!\!\text{ /T)  }\!\!\times\!\!\text{ T)} \displaystyle =\text{ }sin\text{ }4\pi =\text{ }0;\text{ }and\text{ }also\text{ }at\text{ t}=\text{ }0,\text{ }sin\text{ }2\omega t=\text{ }sin\text{ }0\text{ }=\text{ }0].

hence        \displaystyle I_{{rms}}^{2}RT=\frac{{I_{0}^{2}RT}}{2}       or          \displaystyle I_{{rms}}^{2}=\frac{{I_{0}^{2}}}{2}

or      \displaystyle I_{{rms}}^{{}}=\frac{{{{I}_{0}}}}{{\sqrt{2}}} Similarly \displaystyle {{E}_{{rms}}}=\frac{{{{E}_{0}}}}{{\sqrt{2}}}

Illustration

The electric current in a circuit is given \displaystyle i={{i}_{o}}\frac{t}{\tau}for some time. Calculate the rms current for the period t = 0 to \displaystyle t\text{ }=\tau .

Solution

\displaystyle {{I}_{{rms}}}=\sqrt{{\frac{{\int\limits_{0}^{\tau }{{{{I}^{2}}.dt}}}}{{\int\limits_{0}^{\tau }{{dt}}}}}}=\,\,\sqrt{{\frac{{\int\limits_{0}^{\tau }{{\frac{{I_{0}^{2}\,\,{{t}^{2}}}}{{{{\tau }^{2}}}}\,\,.\,\,dt}}}}{\tau }}}

          \displaystyle =\sqrt{{\frac{{I_{0}^{2}}}{{{{\tau }^{3}}}}\left[ {\frac{{{{t}^{3}}}}{3}} \right]_{0}^{\tau }}}=\sqrt{{\frac{{I_{0}^{2}}}{{{{\tau }^{3}}}}.\frac{{{{\tau }^{3}}}}{3}}}-0 = \displaystyle \frac{{{{I}_{0}}}}{{\sqrt{3}}} Ans.

Simple AC Circuits

AC circuit containing only a Resistor

Figure shows a circuit containing an AC source \displaystyle E\text{ }=\text{ }{{E}_{o}}sin\omega t and a resistor of a resistance R.

          Such a circuit is also known as a purely resistive circuit.