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JEE Main 20264 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2026 · 4 April · Morning Shift · Physics · Alternating Current

An inductor of 10 mH, capacitor of 0.1\ μ F and a resistor of 100\...

An inductor of 1010 mH, capacitor of 0.1 μ0.1\ \muF and a resistor of 100 Ω100\ \Omega are connected in series across an a.ca.c power supply 220220 V, 7070 Hz. The power factor of the given circuit is 0.50.5. The difference in the inductive reactance and capacitance reactance is 3α Ω\sqrt{3}\alpha\ \Omega. The value of α\alpha is _____.

Correct answer

0

Step-by-step solution

The power factor of an LCR series circuit is given by cos⁡ϕ=RZ\cos \phi = \dfrac{R}{Z}.

Given cos⁡ϕ=0.5\cos \phi = 0.5 and R=100 ΩR = 100 \ \Omega.

Substituting the values, we get:

0.5=100Z⇒Z=200 Ω0.5 = \dfrac{100}{Z} \Rightarrow Z = 200 \ \Omega

The impedance ZZ of the circuit is also given by:

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Squaring both sides and substituting the values of ZZ and RR:

(200)2=(100)2+(XL−XC)2(200)^2 = (100)^2 + (X_L - X_C)^2

40000=10000+(XL−XC)240000 = 10000 + (X_L - X_C)^2

(XL−XC)2=30000(X_L - X_C)^2 = 30000

∣XL−XC∣=1003 Ω|X_L - X_C| = 100\sqrt{3} \ \Omega

The difference in inductive and capacitive reactance is given as 3α Ω\sqrt{3}\alpha \ \Omega.

Comparing the two expressions:

3α=1003\sqrt{3}\alpha = 100\sqrt{3}

α=100\alpha = 100

Answer: 100100
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