Question

A condenser of $250\mu F$ is connected in parallel to a coil of inductance 0.16 mH, 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)

$8\times {10}^{5}Hz$

For the given case we have standard formula for resonant frequency as:

$f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$

$f=\frac{1}{2\pi }\sqrt{\frac{1}{0.16\times {10}^{-3}\times 250\times {10}^{-6}}-\frac{400}{(0.16\times {10}^{-3}{)}^2}}$

On directly putting the given values in he equation we get:

$\omega =7.98\times {10}^5$