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Solved Problems-11

A coil resonates at 2 MHz when a 18 pF capacitor is shunted across it. When shunting capacitor is 81 pF, the resonating frequency becomes 1 MHz. Find the distributed capacitor of the coil and the self resonating frequency.
Let the distributed capacitance of the coil be Cd (in pF).
Show when the coil is shunted with another capacitance of C, the total capacitance becomes, 
(C + Cd).
When C = 18 pF, the resonating frequency is f0 = 2 MHz
Description: 3862.png(i)
When C = 81 pF, the resonating frequency is f0 = 1 MHz
Description: 3852.png(ii)
Dividing (i) by (ii), we get,
Description: 3868.png
Putting the value of Cd in (ii), we get,
Description: 3874.png
So, the self-resonating frequency of the coil is given as,
Description: 3880.png

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