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MHT CET · Physics · Alternating Current

A series \(\mathrm{L}-\mathrm{C}-\mathrm{R}\) circuit containing a resistance of \(120 \Omega\) has angular frequency \(4 \times 10^5 \mathrm{rad} \mathrm{s}^{-1}\). At resonance the voltage across resistance and inductor are \(60 \mathrm{~V}\) and \(40 \mathrm{~V}\) respectively, then the value of inductance will be

  1. A \(0.2 \mathrm{mH}\)
  2. B \(0.4 \mathrm{mH}\)
  3. C \(0.8 \mathrm{mH}\)
  4. D \(0.6 \mathrm{mH}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.2 \mathrm{mH}\)

Step-by-step Solution

Detailed explanation

At resonance, the impedance of the circuit is equal to resistance
\(
\therefore \mathrm{I}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}=\frac{60}{120}=0.5 \mathrm{~A}
\)
Inductive reactance, \(X_L=\frac{V_L}{I}=\frac{40}{0.5}=80 \Omega\)
\(
\begin{aligned}
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L} \\
& \therefore \mathrm{L}=\frac{\mathrm{X}_{\mathrm{L}}}{\omega}=\frac{80}{4 \times 10^5}=20 \times 10^{-5} \mathrm{H} \\
& =0.2 \mathrm{mH}
\end{aligned}
\)