Q. An $LCR$ 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
A) $\frac{10^{5}}{2\pi}s^{- 1}$
B) $\frac{10^{6}}{2\pi}s^{- 1}$
C) $2\pi \times 10^{5}s^{- 1}$
D) $2\pi \times 10^{6}s^{- 1}$
Explanation
(a) Impedance of $LCR$ circuit will be minimum at resonant frequency so
$V_{0} = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{1 \times 10^{- 3} \times 0.1 \times 10^{- 6}}}$
$= \frac{10^{5}}{2\pi}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.