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A 200Omega resistor , 1.5 H inductorand 35 muFcapacitor are connectedin serieswith a 220V ,50 Hz ac supply.Calculatethe impedanceof the circuitand alsofind the currentthroughthe circuit . |
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Answer» Solution :GIVEN R=20 W ,L=1.5 H , `C=35xx10^(-6)` F `V_(rms)`=220V , f=50 Hz w.k.t `X_L=2pifL` =2 x 3.142 x 50 x 1.5 `=471.3 Omega ~~ 91 Omega` `X_C=1/(2pifC)=1/(2xx3.142xx50xx35xx10^(-6))` `(X_L-X_C)` = 471.3-91 `=380.3 Omega` w.k.t. `Z=sqrt(R^2+ (X_L -X_C)^2)=sqrt(400+144704)` IMPEDANCE, z `~~ 381 Omega` `thereforeI_(rms)=V_(rms)/Z=220/381`=0.578 A `I_(rms)`=r.m.svalue of current= 0.578 A `~~` 0.6 A |
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