The resonant frequency of the L-C circuit is f0 before insertion of the dielectric of - r -16- After inserting the dielectric- the resonant frequency willbeA- 2 f 0B- f 0-4D- 4 f 0

The correct option is B f_04Given, Before introducing the dielectric, the resonant frequency is f_0 = 1/2 π√(LC) After introducing the dielectric, the resonant ...

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Standard XII

Physics

V-I Characteristics for a P-N Junction Diode

The resonant ...

Question

The resonant frequency of the $L - C$ circuit is $f_{0}$ before insertion of the dielectric of $ϵ_{r} = 16$ . After inserting the dielectric, the resonant frequency will be:

\frac{f_{0}}{2}

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2 f_{0}

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\frac{f_{0}}{4}

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4 f_{0}

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Solution

The correct option is C $\frac{f_{0}}{4}$
Given,
Before introducing the dielectric, the resonant frequency is
$f_{0} = \frac{1}{2 π \sqrt{L C}}$
After introducing the dielectric, the resonant frequency is
$f = \frac{1}{2 π \sqrt{L (16 C)}}$ $(C_{n e w} = ϵ_{r} C = 16 C)$
$f = \frac{f_{0}}{4}$

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The resonant frequency of the L−C circuit is f0 before insertion of the dielectric of ϵr=16. After inserting the dielectric, the resonant frequency will be:

The correct option is C f04Given, Before introducing the dielectric, the resonant frequency is f0=12π√LC After introducing the dielectric, the resonant frequency is f=12π√L(16C) (Cnew=ϵrC=16C) f=f04

The resonant frequency of the $L - C$ circuit is $f_{0}$ before insertion of the dielectric of $ϵ_{r} = 16$ . After inserting the dielectric, the resonant frequency will be:

The correct option is C $\frac{f_{0}}{4}$
Given,
Before introducing the dielectric, the resonant frequency is
$f_{0} = \frac{1}{2 π \sqrt{L C}}$
After introducing the dielectric, the resonant frequency is
$f = \frac{1}{2 π \sqrt{L (16 C)}}$ $(C_{n e w} = ϵ_{r} C = 16 C)$
$f = \frac{f_{0}}{4}$