Q103 / 150 · MHT CET 2022 (07 Aug Shift 2) · Physics

At resonance, the value of current in a series L-C-R- (Symbols have their usual meanings.)

  • Correct answer
    A
    $\frac{\mathrm{e}_0}{\mathrm{R}}$
  • B
    $\frac{\mathrm{e}_0}{\sqrt{\mathrm{r}^2+\omega^2 \mathrm{C}^2}}$
  • C
    $\mathrm{e}_0\left[\mathrm{R}^2+\left(\omega \mathrm{L}+\frac{1}{\omega \mathrm{C}}\right)^2\right]$
  • D
    $\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}}$

Solution

At resonance, $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$ So, $\mathrm{i}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2}}=\frac{\mathrm{e}_0}{\mathrm{R}}$

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

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