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At a particular angular frequency, the reactance of capacitor and that of inductor is same. If the angular frequency is doubled, the ratio of the reactance of the capacitor to that of the inductor will be
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Verified Answer
The correct answer is:
$\frac{1}{4}$
We know,
$X_L=\omega L$ ....(i)
$\Rightarrow X_L \propto \omega$
Similarly,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega}$
Given $X_L^1=2 X_L$ $\ldots .(\because \omega=2 \omega)$
$X_C^1=\frac{X_c}{2}$
$\therefore \quad \frac{\mathrm{X}_{\mathrm{C}}^1}{\mathrm{X}_{\mathrm{L}}^1}=\frac{\frac{\mathrm{X}_{\mathrm{C}}}{2}}{2 \mathrm{X}_{\mathrm{L}}}=\frac{1}{4}$
....(from (i))
$X_L=\omega L$ ....(i)
$\Rightarrow X_L \propto \omega$
Similarly,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega}$
Given $X_L^1=2 X_L$ $\ldots .(\because \omega=2 \omega)$
$X_C^1=\frac{X_c}{2}$
$\therefore \quad \frac{\mathrm{X}_{\mathrm{C}}^1}{\mathrm{X}_{\mathrm{L}}^1}=\frac{\frac{\mathrm{X}_{\mathrm{C}}}{2}}{2 \mathrm{X}_{\mathrm{L}}}=\frac{1}{4}$
....(from (i))
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