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

If $L= 10\;mH,C=100\;\mu F,R=100 \;\Omega $ in a LCR circuit the value of resonance frequency is

$(a)\;\frac{10^3}{2 \pi}Hz \\(b)\;2 \times 10^{3} \; Hz \\ (c)\;\frac{2 \times 10^3}{\pi}\;Hz \\ (d)\;10^3\;Hz $

1 Answer

$f= \large\frac{1}{2\pi} \frac{1}{\sqrt {LC}}$
$\qquad= \large\frac{1}{2\pi} \frac{1}{\sqrt {10 \times 10^{-3}\times 100 \times 10^{-6}}}$
$\qquad = \large\frac{10^3}{\pi}$$\;Hz$
Hence a is the correct answer.

 

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

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