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When an inductor tunes at 200 KHz with 624 pF capacitor and
at 600 KHz with 60.4 pF capacitor then the self capacitance
of the inductor would be
a) 8.05 pF b) 10.05pF c.) 16.01pF d.) 20.01pF
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Answer Posted / rs90
the inherent self-capacitance of the inductor coil will be
in parallel with its inductance..so in a tuned circuit this
capacitance will be in parallel with the given external
capacitor..so the resultant capacitance is the sum of these two.
Let the resultant capacitance in the first and 2nd case be
C1 and C2 resp.Since the resonant frequency is inversely
proportional to sqrt(LC) and L being same in both cases..we
can write
C1/C2 = (f2)² /(f1)²
where C1=624 pF C2=60.4 pF f1=200 KHz f2=600 KHz
Solving we get C1/C2=9
now C1=624 + x
C2=60.4 + x
x=self capacitance
put these in above eqn to get x= 10.05
| Is This Answer Correct ? | 10 Yes | 1 No |
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