A circuit consisting of a capacitor and a coil connected in series is fed two alternating voltages of equal amplitudes but different frequencies. The frequency of one voltage is equal to the natural oscillation frequency $(ω_{0})$ of the circuit, the frequency of the other voltage is $η$ times higher. Find the ratio of the current amplitudes. Given quality factor = Q.

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$\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 + \frac{Q^{2}}{η^{2}} {(η^{2} - 1)}^{2}}$

$\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 - \frac{Q^{2}}{η^{2}} {(η^{2} - 1)}^{2}}$

$\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 + \frac{Q^{2}}{η^{2}} {(η^{2} + 1)}^{2}}$

$\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 + \frac{Q^{2}}{η^{2}} {(η^{2} - 1)}^{}}$

answer is A.

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Detailed Solution

At resonance,

$\begin{array}{l} ∴ & ω_{0} = \frac{1}{\sqrt{L C}} \\ ∴ & I_{M_{0}} = \frac{V_{M}}{R} \end{array}$
Also, for second voltage,
$I_{M} = \frac{V_{M}}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$
According to problem, $ω = η ω_{0}$
$∴ I_{M} = \frac{V_{M}}{\sqrt{R^{2} + {(η ω_{0} L - \frac{1}{η ω_{0} C})}^{2}}}$
or $I_{M} = \frac{V_{M}}{\sqrt{R^{2} + {(\frac{η L}{\sqrt{L C}} - \frac{\sqrt{L C}}{η L})}^{2}}}$
or $\frac{I_{M_{0}}}{I_{M}} = \frac{V_{M} / R}{\sqrt{R^{2} + {(\frac{η L}{\sqrt{L C}} - \frac{\sqrt{L C}}{η L})}^{2}}}$
or $\frac{I_{M_{0}}}{I_{M}} = \frac{\sqrt{R^{2} + {(\frac{η L}{\sqrt{L C}} - \frac{\sqrt{L C}}{η L})}^{2}}}{R}$
or $\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 + \frac{1}{R^{2}} {(\frac{η L}{\sqrt{L C}} - \frac{\sqrt{L C}}{η L})}^{2}}$
$\frac{I_{M_{0}}}{I_{M}} = \sqrt{1 + \frac{L C}{R^{2} C^{2} η^{2}} {(η^{2} - 1)}^{2}} = \sqrt{1 + \frac{Q^{2}}{η^{2}} {(η^{2} - 1)}^{2}}$