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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Alternating Current

Resonance frequency of a circuit is f. If capacitance is made 4 time the initial value, the resonance frequency will be

$(a)\;\text{halfed} \\(b)\;\text{double} \\ (c)\;\text{will remain same} \\ (d)\;\text{will be three lines} $

1 Answer

$f= \large\frac{1}{2 \pi \sqrt {LC}}$
$f \alpha \large\frac{1}{\sqrt C}$
$\therefore f$ will become $\large\frac{1}{\sqrt 4}\;$ time
ie f will become $\large\frac{1}{2}$ times
Hence a is the correct answer.

 

answered Mar 27, 2014 by meena.p
edited Sep 24, 2014 by thagee.vedartham
 

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