A series LCR circuit with L=\frac100\pi mH, C=\frac10^-3\pi F and R=10 \Omega , is connected across an ac source of 220 ~V, 50 ~Hz supply. The power factor of the circuit would be \underline\hspace2cm .

Alternating Current PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q82 with Solution

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Correct answer: 1

Given data

$L=\frac{100}{\pi}\text{ mH}=\frac{100}{\pi}\times 10^{-3}\text{ H}=\frac{0.1}{\pi}\text{ H}$

$C=\frac{10^{-3}}{\pi}\text{ F}$

$R=10\,\Omega$

Supply frequency: $f=50\text{ Hz}$
Angular frequency

$\omega=2\pi f=2\pi(50)=100\pi\text{ rad/s}$
Inductive reactance

$X_L=\omega L=(100\pi)\left(\frac{0.1}{\pi}\right)=10\,\Omega$
Capacitive reactance

$X_C=\frac{1}{\omega C}$

Substitute values:

$\omega C=(100\pi)\left(\frac{10^{-3}}{\pi}\right)=0.1$

Hence,

$X_C=\frac{1}{0.1}=10\,\Omega$
Net reactance

$X=X_L-X_C=10-10=0$

So the circuit is in resonance.
Impedance of series LCR circuit

$Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{10^2+0^2}=10\,\Omega$
Power factor

Power factor is $\cos\phi=\frac{R}{Z}=\frac{10}{10}=1$

Therefore, the power factor of the circuit is:

$\boxed{1}$