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 answerA$\frac{\mathrm{e}_0}{\mathrm{R}}$
- 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'.