In a series `L_R` circuit `X_L = R` and power factor of the circuit is `P_1` . When capacitor with capacitance C such that `X_L = X_C` is put in series, the power factor becomes `P_2` . Calculate `P_1/P_2`

(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)

The power factor of L-R circuit is-

In L-C-R series AC circuit,

In a series resonant L-C-R circuit, the power factor is

In L-C-R series circuit

In a series LCR circuit shown below the power factor is unity . Calculate the capacitance C of the capacitor.

The power factor of the circuit shown is p. Now a capacitor ( X_c = 2R ) is also joined in series. Find p_1/p_2