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Alternating Current

JEE Main level, numerical answer, medium

55 of 190

Question 55 of 190

An inductor of $10$ mH, capacitor of $0.1\ \mu$ F and a resistor of $100\ \Omega$ are connected in series across an $a.c$ power supply $220$ V, $70$ Hz. The power factor of the given circuit is $0.5$ . The difference in the inductive reactance and capacitance reactance is $\sqrt{3}\alpha\ \Omega$ . The value of $\alpha$ is _____.

Show answer and solution

Answer: 100

The power factor of an LCR series circuit is given by

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

.

Given

\cos \phi = 0.5

and

R = 100 \ \Omega

.

Substituting the values, we get:

0.5 = \dfrac{100}{Z} \Rightarrow Z = 200 \ \Omega

The impedance

Z

of the circuit is also given by:

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

Squaring both sides and substituting the values of

Z

and

R

(200)^2 = (100)^2 + (X_L - X_C)^2

40000 = 10000 + (X_L - X_C)^2

(X_L - X_C)^2 = 30000

|X_L - X_C| = 100\sqrt{3} \ \Omega

The difference in inductive and capacitive reactance is given as

\sqrt{3}\alpha \ \Omega

.

Comparing the two expressions:

\sqrt{3}\alpha = 100\sqrt{3}

\alpha = 100

Answer:

100

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JEE Main 2025 Apr 2 Shift 1 question paper

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