Question

A condenser of $250\mu F$ is connected in parallel to a coil of inductance $0.16mH$, while its effective resistance is $20\Omega$. Determine the resonant frequency.

• A. $9\times 10^{4}Hz$
• B. $16\times 10^{7}Hz$
• C. $8\times 10^{5}Hz$
• D. $9\times 10^{3}Hz$

Answer 1

Answer: (C)

The resonant frequency is given by
$f_0=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}},$
where $R=$ resistance and
$L=$ inductance.
$\Rightarrow f_0=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$
$=\frac{1}{2\times 3.14}\sqrt{\frac{1}{250\times 10^{-6}\times 0.16\times 10^{-3}}-\frac{20\times 20}{{\left(0.16\times 10^{-3}\right)}^2}}$
$=8\times 10^{5}Hz$