(B) In an $LR$ circuit,the impedance $Z$ is given by $Z = \sqrt{X_{L}^{2} + R^{2}}$.Given $X_{L} = R$,we have $Z = \sqrt{R^{2} + R^{2}} = \sqrt{2}R$.T

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(B) In an $LR$ circuit,the impedance $Z$ is given by $Z = \sqrt{X_{L}^{2} + R^{2}}$.Given $X_{L} = R$,we have $Z = \sqrt{R^{2} + R^{2}} = \sqrt{2}R$.The power factor $P_{1} = \cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}}$.When a capacitor is added such that $X_{L} = X_{C}$,the circuit becomes a series $LCR$ circuit at resonance.In resonance,the impedance $Z = R$.The power factor $P_{2} = \cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$.Therefore,the ratio $\frac{P_{1}}{P_{2}} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.

Class 12 Physics Alternating Current Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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In a series $LR$ circuit,$X_{L} = R$ and the power factor of the circuit is $P_{1}$. When a capacitor with capacitance $C$ such that $X_{L} = X_{C}$ is put in series,the power factor becomes $P_{2}$. The ratio $\frac{P_{1}}{P_{2}}$ is

JEE MAIN 2022 All JEE MAIN

A
$\frac{1}{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{\sqrt{3}}{\sqrt{2}}$
D
$1$

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Alternating Current

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Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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In a series $LR$ circuit,$X_{L} = R$ and the power factor of the circuit is $P_{1}$. When a capacitor with capacitance $C$ such that $X_{L} = X_{C}$ is put in series,the power factor becomes $P_{2}$. The ratio $\frac{P_{1}}{P_{2}}$ is

Similar Questions

$A$ series $LCR$ circuit with $R = 20 \ \Omega$, $L = 1.6 \ \text{H}$ and $C = 40 \ \mu\text{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is . . . . . . $\Omega$.

With the gradual increase in frequency of an $A.C.$ source,the impedance of an $LCR$ series circuit

In a series resonant $LCR$ circuit,for the power dissipated to become half of the maximum power dissipated,the current amplitude is

$A$ resistor of resistance $R$, an inductor of inductive reactance $2R$, and a capacitor of capacitive reactance $X_C$ are connected in series to an $A.C.$ source. If the series $LCR$ circuit is in resonance, then the power factor of the circuit and the value $X_C$ are respectively:

$A$ series $LCR$ circuit with $L = \frac{100}{\pi} \text{ mH}$,$C = \frac{10^{-3}}{\pi} \text{ F}$,and $R = 10 \ \Omega$ is connected across an $AC$ source of $220 \text{ V}, 50 \text{ Hz}$ supply. The power factor of the circuit would be . . . . . . .

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