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An ideal inductor of 5/(pi) H inductance is connected to a 200 V, 50 Hz a.c. supply. (a) Calculate the rms and peak value of current in the inductor. (b) What is the phase difference between current through the inductor and the applied voltage ? How will it change if a small resistance is connected in series with this inductor in the circuit ? |
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Answer» Solution :Here L=`5/piH,V_(rms)=200V` and frequency of a.c. V = 50 Hz (a) The nns value of current `I_(rms)=(V_(rms))/(X_(L))=(V_(rms))/(L.2piv)=200/(5/pixx2pixx50)=0.4A` and peak value of current `I_(m)=sqrt2I_(rms)=sqrt2xx0.4=0.57A` (b) The phase difference between current I through the conductor and the applied voltage is `pi/2` radian and I lags behind V. If a small RESISTANCE R is connected in SERIES with this inductor L, then current I lags behind the voltage V m phase by an phase angle `phi` where `tanphi=(X_(L))/R=(Lomega)/R`. Thus, `0^(@)ltphiltpi/2`. |
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