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Define inductive and capacitive reactance. Give their units. |
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Answer» Solution :Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source (Figure). The alternating voltage is given by the equation. `v=V_(m)sinomegat""...(1)` ii. The alternting current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by Back emf, `epsi=-(di)/(dt)` By applying Kirchoff's loop to the purely inductive circuit, we get `v+epsi=0` `V_(m)sinomegat=L(di)/(dt)` `di=(V_(m))/(L)sinomegatdt` Integrating both sides, we get `i=(V_(m))/(Lomega)(-cosomegat)+"constant"`  iii. The INTEGRATION constant in the above eqution is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero. `cosomega=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 (1) and (2), it is 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 also, it is seen that current lags the voltage by `90^(@)` (Figure). v. Iductive reactance `X_(L)` The peak value of current `I_(m)` is given by `I_(m)=(V_(m))/(omegaL).` Let us 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 reatance `(X_(l)).` It is measured in ohm. `X_(L)=omegaL` AC circuit containing only a capacitor i. Consider a circuit containing only a capacitor of capacitance C connected across an alternating voltage is given by `v=V_(m)sinomegat`  ii. Let q be the instantaneous charge on the capacitor. The emf across the capacitor at that INSTANT is `(q)/(C)` According to Kirchoff's loop rule, `v=(q)/(C)=0` `q=CV_(m)sinomegat` By the definition of current, `i=(dq)/(dt)=(d)/(dt)(CV_(m)sinomegat)` `=CV_(m)(d)/(dt)sinomegat` `ori=(V_(m))/((1)/(Comega))sin(omegat+(pi)/(2))` Instantaneous value of current, `i=I_(m)sin(omegat+(pi)/(2))""...(2)` iii. where `(V_(m))/((1)/(Comega))=I_(m)` the peak value of the alternating current. From equations (1) and (2), it is clear that curret leads the applied voltage by `(pi)/(2)` in a capacitive circuit. This is shown pictorially in Figure. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by `90^(@).` Capacitive reactance `X_(C)` The peak value of current `I_(m)` is given by `I_(m)=(V_(m))/((1)/(Comega)).` Let us compare this equation with `I_(m)=(V_(m))/(R)` from resistive circuit. The quantity `(1)/(Comega)` plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance `(X_(C)).` It measured in ohm. `X_(C)=(1)/(Comega).` |
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