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A circuit containing an `80mH` inductor and a `60(mu)F` capacitor in sereis is connected to a `230V-50Hz` Supply. The resistance in the circuit is negligible . (a) Obtain the current amplitude and rms currents. (b) Obtain the rms values of voltage across inductor and capacitor. (c ) What is the average power transferred to the inductor and to hte capacitor? (d) What is the total power absorbed by the circuit? |
Answer» (a) Given `L=80MH, C=60(mu)F, E_(v)=230V, f=50Hz` Since the circuit has no resistance, so impedance of the circuit is equal to its reactance. `Z=X=X_(C)_X_(L)=(1)/(100 pi C)-100pi L=28 (Omega)` Current amplitude is `I_(0)=(E_0)/(Z)=(230 sqrt(2))/(28)=11.6A` (b) The rms volatge across capacitor is `V_(Lv)=I_(v)X_(L)=(111.6//sqrt(2))100 (pi)L=206 V` (c ) Average power transferred to inductor and capacitor will be zero because whatever energy is used in building up of charges on capacitor (current in case of inductor ) is returned in discharging the capacitor (decay of current in case of inductor). (d) As there is no resistance in the circuit, power factor of hte circuit is `cos(phi) =R//Z=0`. Hence, no power is absorbed by the circuit. |
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