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The maximum capacitance of a variable capacitor is 33pF. What should be the self-inductance to be connected to this capacitor for the natural frequency of the LC circuit to be 810KHz. Corresponding to A.M. broadcast band of Radio Nepal?


Given, Capacitance (C)=33pF=331012F

Let, L be the self-inductance connected with the capacitance, so that the natural frequency of this LC circuit is 810KHz,

i.ef=810103Hz

Now, we have 

f=12πLC

L= 14π2(810103)2331012  =1.17103H

Thus, Inductance connected (L)=1.17103H
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