Question

In a Q-meter measurement to determine self capacitance of coil, the first resonance occurs at $f_1$ with $C_1=1000pF$ and the second resonance occurs at $f_2=3f_1$ with $C_2=100pF.$ The self capacitance of the coil is

• A. 12.5 pF
• B. 21.5 pF
• C. 31.2 pF
• D. 19.3 pF

Answer 1

The correct option is A 12.5 pF

(a) For measuring self/distributed capacitance



For first resonance, $f_1=\frac{1}{2\pi \sqrt{L(C_1+C_d)}}$

For second resonance, $f_2=\frac{1}{2\pi \sqrt{L(C_2+C_d)}}$

and $f_2=n{f}_1$

Equating we get,
Self/ distributed capacitance,

$C_d=\frac{C_1-n^2{C}_2}{n^2-1}=\frac{1000-9\left(100\right)}{9-1}$ $(asn=3)$
$=\frac{1000-900}{8}=\frac{100}{8}=12.5pF$