Q104 / 150 · MHT CET 2021 (24 Sep Shift 1) · Physics

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

  • Correct answer
    A
    $0.2 \mathrm{mH}$
  • B
    $0.4 \mathrm{mH}$
  • C
    $0.8 \mathrm{mH}$
  • D
    $0.6 \mathrm{mH}$

Solution

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, $X_L=\frac{V_L}{I}=\frac{40}{0.5}=80 \Omega$ $$ \begin{aligned} & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L} \\ & \therefore \mathrm{L}=\frac{\mathrm{X}_{\mathrm{L}}}{\omega}=\frac{80}{4 \times 10^5}=20 \times 10^{-5} \mathrm{H} \\ & =0.2 \mathrm{mH} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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