A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80 \Omega , an inductor of inductive reactance X_L=70\Omega , and a capacitor of capacitive reactance X_C=130\Omega . The power factor of circuit is \fracx10 . The value of x is :

Alternating Current PYQ — JEE Main 2023 · 25 Jan · Shift 2 · Q70 with Solution

View written solutionFree

Correct answer: 8

We need the power factor of the series LCR circuit.

For a series LCR circuit,

X = X_L - X_C = 70-130 = -60\,\Omega

So the circuit is overall capacitive, but for impedance magnitude we use $|X|=60\,\Omega$ .

The impedance is

Z=\sqrt{R^2+(X_L-X_C)^2}

Substitute the values:

Z=\sqrt{80^2+(-60)^2}

Z=\sqrt{6400+3600} = \sqrt{10000}=100\,\Omega

For a series AC circuit,

\cos\phi = \frac{R}{Z}

Thus,

\cos\phi = \frac{80}{100}=0.8

Given that power factor is $\frac{x}{10}$ ,

\frac{x}{10}=0.8=\frac{8}{10}

Hence,

x=8

Our derived answer is $8$ , which matches the stored correct answer.