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Obtain an equation of current for AC voltage applied to an inductor and draw a graph of V and I. |
Answer» Solution :In figure shows an AC source connected to an inductor.  Inductor has negligible resistance. Thus, the CIRCUIT is a purely INDUCTIVE AC circuit. Let the voltage across the source be `V= V_(m) sin omega t ` . Using the Kirchoff.s loop rule, `V - L (dI)/(dt ) = 0 ` where `- L (dI )/(dt)` is the self induced emf. `:. V = L ( dI )/( dt ) ` ` :. (dI)/(dt) = ( V )/( L )` But `V = V_(m) sin omega t ` `:. ( dI )/( dt ) = ( V_(m) sin omega t )/( L )` `:. dI = ( V _(m))/( L ) sin omega dt `.....(1) where L is the self inductance. Equation (1) indicates that current I ( t) as a function of time,must be such that its slope `(dI)/(dt)` is sine an sinusoidally varying quantity with the same phase as the source voltage and an amplitude given by `(V_(m))/( L )`. To obtain the current integrate equations (1) with respective to time, `:. int dI = (V_(m))/( L ) int sin omega t dt ` `:. I = ( V_(m))/( L )XX ( cos omega t )/( omega ) +` constant....(2) Here, integration constant has the dimension of current and is time INDEPENDENT At t=0, time I = 0 `:.` 0 `+` constant `:.` Constant =0 `:.` From equation (2), `:. I = - (V_(m))/( L omega ) cos omega t ` `:. I = ( V_(m))/( omega L ) sin ( omega t - ( pi )/( 2)) [ :. - cos omega t = sin ( omega t - ( pi )/( 2))]` `omega L` is called inductive reactance denoted by `X_(L)`. `:. X_(l) = omega L ` and its unit is Ohm. The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistivecircuit. The inductive reactance is directly proportional to the inductance and frequency of the current. Hence , for the source voltage and the current in an indeuctor `V = V_(m) sin omega t ` and `I = I _(m) sin( omega t - ( pi )/( 2))` . This shows the that hte current lags the voltag by `( pi )/(2)` or one -quarter `((1)/(4))` cycle. In this case figure (a) shows that voltage and current phasors in the presentcase at instant `( t_(1))`.  The current PHASOR I is `( pi )/(2)` behind the voltage phasor V. ( Lags) `( pi )/(2) = ( T )/( 4):. ( pi )/( 2) = ( T )/( 2pi )xx ( 2pi )/( 4) = (( pi )/( 2))/( omega ) = ( pi )/( 2 omega )` When rotated with frequency `omega ` counter clockwise, they generate the voltage and current given by `V = V_(m) sin omega t ` and `I = I _(m) sin ( omega t - ( pi )/( 2))` respectively and as shown by figure (b). This graphs is for V and I versus `omega t `. |
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