Question

What is the power factor of LCR circuits at resonance?

Answer 1

Hint : Power factor is the ratio of active power and apparent power. Reactive power of LCR combination is the difference between capacitive and inductive power. At resonance, capacitive and inductive reactances are equal.

Formula Used: The formulae used in the solution are given here.
 $\Rightarrow P.F. = \dfrac{R}{Z} $ where $ P.F. $ is the power factor, $ R $ is the reactance and $ Z $ is the total impedance.
 $\Rightarrow Z = \sqrt {{R^2} + \left( {{X_C} - {X_L}} \right)} $ where , $ R $ is the reactance and $ Z $ is the total impedance and $ {X_C} $ is the capacitive reactance and $ {X_L} $ is the inductive reactance.
 $\Rightarrow {X_C} = \dfrac{1}{{\omega C}} $ where $ C $ is the capacitance and $ {X_L} = \omega L $ where $ L $ is the inductance.
 $\Rightarrow \omega = 2\pi f $ where $ f $ is the frequency of AC source.

Complete step by step answer
Power factor is also the ratio of resistance of LCR circuit to its impedance. The power factor of an LCR circuit is the ratio of the resistance to the total impedance of the circuit. The total impedance consists of the magnitude of the phasor sum of the resistance, the capacitive reactance and the inductive reactance.
The power factor is a measure of the fraction of total power that is being used up or dissipated by a load resistor. Since, capacitors and inductors do not dissipate power but keep on exchanging them between the source and themselves, they do not utilize the power. The higher the power factor, the better it is for energy efficiency as a greater fraction of the power is available for utilization.
We know that, $ P.F. = \dfrac{R}{Z} $ where $ P.F. $ is the power factor, $ R $ is the reactance and $ Z $ is the total impedance.
 $\Rightarrow Z = \sqrt {{R^2} + \left( {{X_C} - {X_L}} \right)} $ where $ R $ is the reactance and $ Z $ is the total impedance and $ {X_C} $ is the capacitive reactance and $ {X_L} $ is the inductive reactance.
At resonance, $ {X_C} = {X_L} $ .
We know, $ {X_C} = \dfrac{1}{{\omega C}} $ where $ C $ is the capacitance and $ {X_L} = \omega L $ where $ L $ is the inductance.
 $\Rightarrow \omega = 2\pi f $ where $ f $ is the frequency of AC source.
 $\Rightarrow \dfrac{1}{{\omega C}} = \omega L $
 $ \Rightarrow \omega = \sqrt {\dfrac{1}{{LC}}} $
Since, $ {X_C} = {X_L} $ , thus $ {X_C} - {X_L} = 0 $ .
Total impedance $ Z $ is thus, $ Z = \sqrt {{R^2} + 0} $
 $ \Rightarrow Z = R. $
To find the power factor,
 $\Rightarrow P.F. = \dfrac{R}{Z} = 1 $ .
$ \therefore $ The power factor at resonance is 1.

Note
The voltage across the inductor and capacitor cancel each other. The total impedance of the circuit is only due to the resistor. The phase angle between voltage and current is zero.
 $\Rightarrow \cos \varphi = \dfrac{R}{Z} = \dfrac{R}{R} = 1. $ The power factor is unity.