Question

A Q-meter having an insertion resistance of $0.02\Omega$ is used to measure the inductance of a coil. Resonance occurs at an angular frequency of ${10}^6$ rad/sec with a capacitance of 40 pF. The inductance of the coil is

• A. $250\mu H$
• B. 2.5 mH
• C. 25 mH
• D. 0.25 H

Answer 1

The correct option is C 25 mH

At resonance,

${\omega }_c=\frac{1}{\sqrt{LC}}$

$\therefore {10}^6=\frac{1}{\sqrt{L\times 40\times {10}^{-12}}}$

$({10}^6{)}^2=\frac{1}{L\times 40\times {10}^{-12}}$

$L=\frac{1}{40\times {10}^{-12}\times {10}^{12}}$

$=\frac{1}{40}=0.025H=25mH.$