An L-C-R circuit contains R=50 Ω, L=1 mH and C=0.1 μ F . The impedence of the circuit will be minimum for a frequency of

Q119 / 120 · VITEEE 2014 · Physics

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

Correct answer
A
$\frac{10^{5}}{2 \pi} \mathrm{Hz}$
B
$\frac{10^{6}}{2 \pi} \mathrm{Hz}$
C
$2 \pi \times 10^{5} \mathrm{~Hz}$
D
$2 \pi \times 10^{6} \mathrm{~Hz}$

Solution

Impedance of L-C-R circuit will be minimum for a resonant frequency so, $\begin{aligned} & v_{0}=\frac{1}{2 \pi \sqrt{L C}} \\ =& \frac{1}{2 \pi \sqrt{1 \times 10^{-3} \times 0.1 \times 10^{-6}}}=\frac{10^{5}}{2 \pi} \mathrm{Hz} \end{aligned}$

Hence, the correct answer is option: 'A'.

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