Answer
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Hint: 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. At resonance, the capacitive and inductive reactances are equal.
Formula used:
$\text{Power factor }\left( p.f \right)=\dfrac{\text{Resistance }\left( R \right)}{\text{Total impedance (}Z\text{)}}$
$\text{Total Impedance (}Z\text{) = }\sqrt{\left( {{R}^{2}} \right)+{{\left( {{X}_{c}}-{{X}_{L}} \right)}^{2}}}$
where ${{X}_{c}}=\text{ Capacitive reactance = }\dfrac{1}{\omega C}$
${{X}_{L}}=\text{ Inductive reactance = }\omega \text{L}$,
where $\omega =2\pi f$and $f=\text{ frequency of AC source}$ and C and L are the capacitance and the inductance in the circuit respectively.
Complete step-by-step answer:
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.
$\text{Power factor }\left( p.f \right)=\dfrac{\text{Resistance }\left( R \right)}{\text{Total impedance (}Z\text{)}}$
$\text{Total Impedance (}Z\text{) = }\sqrt{\left( {{R}^{2}} \right)+{{\left( {{X}_{c}}-{{X}_{L}} \right)}^{2}}}$
$\therefore pf=\dfrac{R}{\sqrt{{{R}^{2}}+{{\left( {{X}_{C}}-{{X}_{L}} \right)}^{2}}}}$ ---(1)
Where ${{X}_{c}}=\text{ Capacitive reactance = }\dfrac{1}{\omega C}$
${{X}_{L}}=\text{ Inductive reactance = }\omega \text{L}$,
where $\omega =2\pi f$and $f=\text{ frequency of AC source}$ and C and L are the capacitance and the inductance in the circuit respectively.
At resonance, the frequency is such that ${{X}_{C}}={{X}_{L}}$
$\therefore \dfrac{1}{\omega C}=L\omega $
$\therefore \omega =\dfrac{1}{\sqrt{LC}}$
${{X}_{C}}-{{X}_{L}}=0$ --(2)
Using (1) and (2),
$pf=\dfrac{R}{\sqrt{{{R}^{2}}+0}}=\dfrac{R}{\sqrt{{{R}^{2}}}}=\dfrac{R}{R}=1$
Therefore, the power factor of an LCR circuit at resonance is 1.
Note: At resonance, the LCR circuit behaves like a purely resistive circuit and the effects of the capacitor and inductor cancel each other out. This is the most efficient circuit for operation. Thus, many electric companies even give incentives to commercial entities if they have a power factor very close to 1.
Formula used:
$\text{Power factor }\left( p.f \right)=\dfrac{\text{Resistance }\left( R \right)}{\text{Total impedance (}Z\text{)}}$
$\text{Total Impedance (}Z\text{) = }\sqrt{\left( {{R}^{2}} \right)+{{\left( {{X}_{c}}-{{X}_{L}} \right)}^{2}}}$
where ${{X}_{c}}=\text{ Capacitive reactance = }\dfrac{1}{\omega C}$
${{X}_{L}}=\text{ Inductive reactance = }\omega \text{L}$,
where $\omega =2\pi f$and $f=\text{ frequency of AC source}$ and C and L are the capacitance and the inductance in the circuit respectively.
Complete step-by-step answer:
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.
$\text{Power factor }\left( p.f \right)=\dfrac{\text{Resistance }\left( R \right)}{\text{Total impedance (}Z\text{)}}$
$\text{Total Impedance (}Z\text{) = }\sqrt{\left( {{R}^{2}} \right)+{{\left( {{X}_{c}}-{{X}_{L}} \right)}^{2}}}$
$\therefore pf=\dfrac{R}{\sqrt{{{R}^{2}}+{{\left( {{X}_{C}}-{{X}_{L}} \right)}^{2}}}}$ ---(1)
Where ${{X}_{c}}=\text{ Capacitive reactance = }\dfrac{1}{\omega C}$
${{X}_{L}}=\text{ Inductive reactance = }\omega \text{L}$,
where $\omega =2\pi f$and $f=\text{ frequency of AC source}$ and C and L are the capacitance and the inductance in the circuit respectively.
At resonance, the frequency is such that ${{X}_{C}}={{X}_{L}}$
$\therefore \dfrac{1}{\omega C}=L\omega $
$\therefore \omega =\dfrac{1}{\sqrt{LC}}$
${{X}_{C}}-{{X}_{L}}=0$ --(2)
Using (1) and (2),
$pf=\dfrac{R}{\sqrt{{{R}^{2}}+0}}=\dfrac{R}{\sqrt{{{R}^{2}}}}=\dfrac{R}{R}=1$
Therefore, the power factor of an LCR circuit at resonance is 1.
Note: At resonance, the LCR circuit behaves like a purely resistive circuit and the effects of the capacitor and inductor cancel each other out. This is the most efficient circuit for operation. Thus, many electric companies even give incentives to commercial entities if they have a power factor very close to 1.
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