Question

(i) An a.c. Source of voltage `V = V_(0) " sin " omegat` is connected to a series combination of L,C and R . Use the phasor diagram to obtain expression for impedance of the circuit and phase angle between voltage and current . Find the condition when current will be phase with the voltage . What is the circuit in this condition called ?
(ii) In a series LR circuit `X_(L) = R` and power factor of the circuit is `P_(1)`. When capacitor which capacitance C such that `X_(L) =X_(C)` is put in series , the power factor becomes `P_(2)` .Calculate `P_(1)//P_(2)`

Answer 1

(i) from the phasor diagram `V^(2) =V_(4)^(2) + (V_(L) -V_(C ))^(2)`
`rArr V = sqrt(V_(R )^(2)+(V_( C) -V_(C ))^(2))`
` V_( R ) =R I , V_( C ) =X_( C ) I , V_(L ) = X_(L ) I`
`rArr V = sqrt(R^(2) I^(2) +(X_(C ) -X_(L))^(2)I^(2))=I sqrt(R^(2) (X_( C) -X_(L)))^(2)`
` (V)/(I)= sqrt(R^(2)(X_(C ) -X_(L))^(2)) rArr Z = sqrt(R^(2)( X_(C ) - X_(L)))^(2)`
This is required impedance of the circuit .
Now ` " tan " phi = (V_(C ) -V_(L))/(V_(R )) =(IX_(C ) -I X_(L))/(IR)`
`" tan " phi = (X_(C ) -X_( L))/(R )`
When `X_( C) = X_(L)` impedance of circuit beomes minimum and current becomes maximum this condition is called condition of resonance
`X_(C ) - X_(L) " " rArr omegaC = 1//omegaL rArr omega^(2) = (1)/(LC ) rArr omega = sqrt((1)/(LC))`
this is resonance fruquency
(ii) For LR series circuit . ,
Power factor `P_(1)=( R)/(Z) = (R )/(sqrt(X_(L)^(2) +R^(2))) =(R )/(sqrt(R^(2) +R^(2))) =(R )/(sqrt(2R^(2))) =(1)/(sqrt(2))`
For LCr series
`P_(2) = (R )/(sqrt((X_(L) -X_(C))^(2)+R^(2)) )=(R )/(sqrt((R-R)^(2) +R^(2)))=1 rArr P_(1): P_(2) =1 : sqrt(2)`