An alternating e.m.f. of $0 \cdot 2 \mathrm{~V}$ is applied across an LCR series circuit having…

An alternating e.m.f. of $0 \cdot 2 \mathrm{~V}$ is applied across an LCR series circuit having $\mathrm{R}=4 \Omega, \mathrm{C}=$ $80 \mu \mathrm{F}$ and $\mathrm{L}=200 \mathrm{mH}$. At resonance the voltage drop across the inductor is
  1. 1V
  2. 2.5V
  3. 3.5V
  4. 10V

Solution

At resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\therefore$ Current $=\frac{0.2}{R}=\frac{0.2}{4}=\frac{0.1}{2}=0.05 \mathrm{~A}$ Now, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}$ $\begin{aligned} \mathrm{f} &=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \\ \therefore \quad \mathrm{X}_{\mathrm{L}} &=\frac{2 \pi \mathrm{L}}{2 \pi \sqrt{\mathrm{LC}}}=\sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \\ \mathrm{X}_{\mathrm{L}} &=\sqrt{\frac{200 \times 10^{-3}}{80 \times 10^{-6}}}=\frac{10^{2}}{2}=50 \Omega \\ \therefore \quad \mathrm{V}_{\mathrm{L}} &=\mathrm{i} \times \mathrm{L}=0.05 \times 50=2.5 \mathrm{~V} \end{aligned}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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