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A series LCR circuit is connected to an a.c. source having voltage V = V_(m) sin omega t. Derive the expression for the instantaneous current I and its phase relationship to the applied voltage. Obtain the condition for resonance to occur. Define 'power factor'. State the conditions under which it is (i) maximum and (ii) minimum. |
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Answer» Solution :From the phasor diagram we have `V_(m)^(2) = V_(Rm)^(2) + (V_(Cm) - V_(Lm))^(2)` `= i_(m)^(2)[R^(2) + (X_(C) - X_(L))^(2)]` `i_(m) = (V_(m))/(sqrt(R^(2) + (X_(C) - X_(L))^(2))`  The current is seen to lead the voltage by an angle `phi` where `tan phi = (X_(C) - X_(L))/(R )` Hence `i = i_(m) sin(OMEGA t + phi)` Where `i_(m) = (V_(m))/(sqrt(R^(2) + (X_(C) - X_(L))^(2))` and `phi = tan^(-) [((OMEGAL -(1)/(omega C)))/(R )]` Condition of resonance : `omega L - (1)/(omega C = 0` or `omega L = (1)/(omega C` or `omega = (1)/(sqrt(LC))` Power factor equals the cosine of the phase angle power factor i.e., power factor, `COS phi = (R )/(Z)` Power FACTORIS maximum when `cos phi = 1` i.e.,when R = Z or `X_(L) = X_(C)`. Power factor is minimum when `cos phi = 0` i.e., when `R = 0`. |
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