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Find out the phase relationship between voltage and current in a pure inductive circuit. |
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Answer» Solution :i. Consider a circuit containing a pure inductor of inductance L CONNECTED across analternating voltage source (figure). Thealternating voltage is given by the equation. `v=V_(m)sinomegat""...(1)` ii. The alternating current FLOWING through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by Back emf,`omega=-(di)/(dt)` By applying Kirchoff's loop to the purely inductive circuit, we get `v+omega=0` `V_(m)sinomegat=L(di)/(dt)` `di=(V_(m))/(L)sinomegadt` Integrating both sides, we get `i=(V_(m))/(L)intsinepsitdt` `i=(V_(m))/(Lomega)(-COSOMEGAT)+"constant"`  III. The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent partk, we can set the time independent part in the current (integration constant) into zero. `cosomegat=sin((pi)/(2)-omegat)` `-sin((pi)/(2)-omegat)=sin(omegat-(pi)/(2))` `i=(V_(m))/(omegaL)sin(omegat-(pi)/(2))` `i=I_(m)sin(omegat-(pi)/(2))""...(2)` iv. where `(V_(m))/(omegaL)=I_(m)` the peak value of the alternating current in the circuit. From equation (1) and (2), itis evident that current lags behind the applied voltage by `(pi)/(2)` in an inductive circuit. This fact is depicted in the phasor diagram. In the wave diagram ALDO, it is seen that current lags the voltage by `90^(@)` (Figure). v. Inductive reactance `X_(L)` The peak value of current `I_(m)` is given by `I_(m)=(V_(m))/(omegaL).` Let us of compare this equation with `I_(m)=(V_(m))/(R)` from resistive circuit The quantity `omegaL` plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance `(X_(L)).` It is measured in ohm. `X_(L)=omegaL` `X_(L)=2piL`  Where f is the frequency of the alternating current. For a steady current, f = 0. Therefore, `X_(L)=0.` Thus an ideal inductor offer no resistance ot steady DC current. |
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