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Showthat voltage in an inductor leads the current by pi//2rad for a pure inductor |
Answer» Solution :  <BR>Consider an inductor of inductance L is connected across an AC source . Let `V= V_(0) sinomegat` …. (1) the self induced emf in the conductor is `epsilon = - L (dI)/(dt)`, ACCORDING to K irchoff loop rule `v- L (dI)/(dt) = 0` `RightarrowV_(0)sinomegat-L(dt)/(dt)=0` `RightarrowL (dt)/(dt)= V_(0)sinomegat.This INDICATES the current in an inuctor is a function of time . `dt= v_(0)/(L)sin omegat` dt To obtain the current at any tan t, we INTEGRATE above equation `I = int di = (V_(0))/(L )int sin omegat dt` `Rightarrow I = (V_(0))/(L)[-cosomegat0/(omega) + constant]` `I = (V_(0))/(L) [(cosomegat0/(omega)+ constant]` `I= (V_(0))/(L_(omega) (-COSOMEGAT)`,br> If we take `(V_(0))/(Lomega)= I_(0)`, the amplitude of the current , then `I = I_(0)(-cosomegat)` `therfore I = I_(0)sin (omegat- (pi)/(2))rightarrow (2)` From (1) and (2) we can conclude that voltage leads current by `(pi)/(2)`,br> Phasor diagram : 
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