MHT CET · Physics · Alternating Current
At resonance, the value of current in a series L-C-R- (Symbols have their usual meanings.)
- 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 \(\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{e}_0}{\mathrm{R}}\)
Step-by-step Solution
Detailed explanation
At resonance,
\(\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}\)
So,
\(\mathrm{i}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2}}=\frac{\mathrm{e}_0}{\mathrm{R}}\)
\(\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}\)
So,
\(\mathrm{i}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}}=\frac{\mathrm{e}_0}{\sqrt{\mathrm{R}^2}}=\frac{\mathrm{e}_0}{\mathrm{R}}\)
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