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At resonance, the value of current in a series L-C-R- (Symbols have their usual meanings.)
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Verified Answer
The correct answer is:
$\frac{\mathrm{e}_0}{\mathrm{R}}$
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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