A series circuit consisting of a capacitor and a coil with active resistance is connected to a source of harmonic voltage whose frequency can be varied, keeping the voltage amplitude constant. At frequencies $ω_{1}$ and $ω_{2}$ , the current amplitudes are n times less than the resonance amplitude. Then resonance frequency will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{ω_{1} + ω_{2}}{2}$

$ω_{1} + ω_{2}$

$ω_{1} - ω_{2}$

$\sqrt{ω_{1} ω_{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I_{M_{1}} = \frac{V_{M}}{\sqrt{R^{2} + {(ω_{1} L - \frac{1}{ω_{1} C})}^{2}}}$
Similarly, $I_{M_{2}} = \frac{V_{M}}{\sqrt{R^{2} + {(ω_{2} L - \frac{1}{ω_{2} C})}^{2}}}$
Since, both $I_{M_{1}} and I_{M_{2}}$ are n times less than resonance current amplitude,
$\begin{array}{l} ∴ I_{M 1} = I_{M 2} \\ ∴ \frac{V_{M}}{\sqrt{R^{2} + {(ω_{1} L - \frac{1}{ω_{1} C})}^{2}}} = \frac{V_{M}}{\sqrt{R^{2} + {(ω_{2} L - \frac{1}{ω_{2} C})}^{2}}} \end{array}$
or $ω_{1} L - \frac{1}{ω_{1} C} = \frac{1}{ω_{2} C} - ω_{2} L$
or $\frac{ω_{1}^{2} LC - 1}{ω_{1} C} = \frac{1 - ω_{2}^{2} LC}{ω_{2} C}$
or $ω_{1} ω_{2} LC (ω_{1} + ω_{2}) - (ω_{1} + ω_{2}) = 0$
or $1 - ω_{1} ω_{2} LC = 0$
$\begin{array}{l} ∴ ω_{1} ω_{2} = \frac{1}{LC} = ω_{0}^{2} \\ ∴ ω_{0} = \sqrt{ω_{1} ω_{2}} \end{array}$