Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1466 Upvotes
Verified Answer
The correct answer is:
$\frac{10^{5}}{2 \pi} \mathrm{Hz}$
Impedance of L-C-R circuit will be minimum for a resonant frequency so,
$\begin{aligned}
& v_{0}=\frac{1}{2 \pi \sqrt{L C}} \\
=& \frac{1}{2 \pi \sqrt{1 \times 10^{-3} \times 0.1 \times 10^{-6}}}=\frac{10^{5}}{2 \pi} \mathrm{Hz}
\end{aligned}$
$\begin{aligned}
& v_{0}=\frac{1}{2 \pi \sqrt{L C}} \\
=& \frac{1}{2 \pi \sqrt{1 \times 10^{-3} \times 0.1 \times 10^{-6}}}=\frac{10^{5}}{2 \pi} \mathrm{Hz}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.