Q75 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
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 _____.
Correct answer: 100
Solution
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: $100$
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: $100$
Hence, the correct answer is 100.