Physics | VITEEE 2014 | Discussion Page

An L-C-R circuit contains $R= 50 \Omega$ , L=1 mH and C= $0.1 \mu F$ . The impedance of the circuit will be minimum for a frequency of

$\frac{10^{5}}{2 \pi}$ Hz

$\frac{10^{6}}{2 \pi}$ Hz

$2 \pi \times 10^{5}$ Hz

$2 \pi \times 10^{6}$ Hz

Option A

Impedence of L-C-R circuit will be minimum for a resonant frequency so,

$v_{0}=\frac{1}{2\pi\sqrt{LC}}$

= $v_{0}=\frac{1}{2\pi\sqrt{1\times10^{-3}\times0.1\times10^{-6}}}=\frac{10^{5}}{2 \pi}Hz$

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