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Applications of Q-Meter

An inductor o...

Question

An inductor of a Q meter tuned to $4 M H z$ with $900 p F$ and to $8 M H z$ with $180 p F$ , the value of self distributed capacitance of the inductor will be _____ $p F .$
60

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Solution

The correct option is A 60
Equivalent of all inter turn capacitances $=$ self distributed capacitance

Resonance frequency

$f_{0} = \frac{1}{2 π \sqrt{L (C + C_{d})}}$

$a t 4 M H z : C_{1} = 900 p F$

$a t 8 M H z : C_{2} = 180 p F$

Now,
$f_{1} = \frac{1}{2 π \sqrt{L (C_{1} + C_{d})}}$

$a n d f_{2} = \frac{1}{2 π \sqrt{L C_{2} + C_{d})}}$

$a n d \frac{f_{2}}{f_{1}} = \sqrt{\frac{C_{1} + C_{d}}{C_{2} + C_{d}}}$

$\frac{8}{4} = \sqrt{\frac{900 + C_{d}}{180 + C_{d}}}$

$4 (180) + 4 C_{d} = 900 + C_{d}$

$3 C_{d} = 900 - 720$

$3 C_{d} = 180$

Distributed capacitance
$C_{d} = 60 p F$

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