Q123 / 150 · MHT CET 2022 (08 Aug Shift 1) · Physics

Resistor of $2 \Omega$, inductor of $100 \mu \mathrm{H}$ and capacitor of $400 \mathrm{pF}$ are connected in series across an a.c. source of $e_{r m s}=0.1$ volt. At resonance, voltage drop across inductor is

  • A
    20V
  • Correct answer
    B
    25 V
  • C
    2.5 V
  • D
    250 V

Solution

We know, at resonance $X_C=X_L$ $\Rightarrow \omega L=\frac{1}{\omega C} \Rightarrow \omega=\left(\frac{1}{\sqrt{L C}}\right)$ The potential drop across inductor is $V_L=i X_L=i \sqrt{\frac{L}{C}}$ $\because$ LCR circuit in series and at resonance $i=\frac{V}{R}=\frac{0.1}{2} \mathrm{~A}=0.05 \mathrm{~A}$ Now, $V_L=0.05 \sqrt{\left(\frac{100 \times 10^{-6}}{400 \times 10^{-12}}\right)} \mathrm{V}=2.5 \times 10^{-2} \times 10^3 \mathrm{~V}=25 \mathrm{~V}$

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

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