Question

A high Q coil having distributed (self) capacitance is tested with a Q-meter. First resonance at ${\omega }_1={10}^6$ rad/s is obtained with a capacitance of 990 pF. The second resonance at ${\omega }_2=2\times {10}^6$ rad/s is obtained with a 240 pF capacitance. The value of the inductance (in mH) of the coil is (up to one decimal place)

• A. 1

Answer 1

The correct option is A 1
Given,
${\omega }_1={10}^6$ $C_1=990pF$

${\omega }_2=2\times {10}^6$ rad/sec $C_2=240pF$

Here,
${\omega }_2=2{\omega }_1$

$\therefore C_d=\frac{C_1-4C_2}{3}$

$=\frac{990-4\times 240}{3}=10pF$

We know at resonance

$f=\frac{1}{2\pi \sqrt{L(C+C_d)}}$

$orL=\frac{1}{4{\pi }^2{f}^2(C+C_d)}$

$TotalC=C_1=990pF$

$L=\frac{1}{{\omega }_1^2(C+C_d)}$

$=\frac{1}{{10}^6\times {10}^6\times (990+10)\times {10}^{-12}}$

$=1.0mH$