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Obtain an expression for the impedance of a series LCR circuit. (using phasor diagram method). |
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Answer» Solution :CONSIDER a resistance R, an inductor of self inductance L and a capacitance C CONNECTED in series across an AC source. The applied voltage is given by, `v=v_(0)sinomegat`  where, v is the instantaneous value, `v_(0)` is the peak value and `omega=2pif`, f being the frequency of AC. If i be the instantaneous current at TIME t, the instantaneous voltages across R, L and C are respectively `iR, iX_(L)andiX_(C)`. The vector sum of the voltage amplitudes across R, L, C EQUALS the amplitude `v_(0)` of the voltage applied. Let `v_(R),v_(L)` and `v_(C)` be the voltage amplitudes across R, L and C respectively and `I_(0)` the current amplitude. Then `v_(R)=i_(0)R` is in phase with `i_(0)`. `v_(L)=i_(o)X_(L)=i_(0)(omegaL)` leads `i_(0)` by `90^(@)` `v_(c)=i_(0)X_(C)=i_(0)((1)/(omegaC))` lags behind `i_(0)` by `90^(@)` The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by `(pi)/(2)` rad. The current in a pure capacitor leads the voltage by `(pi)/(2)` rad. For `v_(L)gtv_(C)`, phase angle `phi` between the voltage and the current is positive. From the RIGHT angled triangle OAP, `OP^(2)=OA^(2)+AP^(2)=OA^(2)+OB^(2)(becauseAP=OB)` `v^(2)=v_(R)^(2)+(v_(L)-v_(C))^(2)=(iR)^(2)+(iX_(L)-iX_(C))^(2)=i^(2)(R^(2)+(X_(L)-X_(C))^(2))` `i=(v)/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))=(v)/(Z)andZ=sqrt(R^(2)+(X_(L)-X_(C))^(2))` Where Z is the impedance of the circuit. Phase angle between v & i. `tanphi=(v_(L)-v_(C))/(v_(R))=(X_(L)-X_(C))/(R_(L)),phi=tan^(-1)((X_(L)-X_(C))/(R))` 
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