JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2026
- Medium
- Numerical
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 _____.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 100
Step-by-step Solution
Detailed explanation
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: \(100\)
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