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`

Find the power factor of the circuit shown in figure.

Find the power factor of the circuit shown in figure.

In series L-R circuit, X_L=R . Now a capacitor with X_C=R is added in series. New power factor

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

In series LR circuit, X_(L) = 3 R . Now a capacitor with X_(C ) = R is added in series. The ratio of new to old power factor

The power factor of an L-R series circuit is 0.5 and that of a C-R series circuit is 0.2. If the elements (L, C and R) of the two circuit are joined in series and connected to the same ac source, the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

Power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5. If the element (L. C, and R) of the two circuits are joined in series the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

(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)