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# A radio can tune over the frequency range of a portion of MW broadcast band:

Q7.10  A radio can tune over the frequency range of a portion of $MW$ broadcast band:  $(800kHz\: to1200kHz)$  . If its $LC$circuit has an effective inductance of $200\mu H$, what must be the range of its variable capacitor?

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Given,

Range of the frequency in which radio can be tune = $(800kHz\: to1200kHz)$

The effective inductance of the Circuit = $200\mu H$

Now, As we know,

$w^2=1/\sqrt{LC}$

$C=1/w^2L$

where $w$ is tuning frequency.

For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum.

first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz.

$C_{minimum}=\frac{1}{w_{minimum}^2L}=\frac{1}{(2\pi(800*10^3))^2*200*10^{-6}}=1.981*10^{-10}F$

Hence the minimum value of capacitance is 198pF.

Now, Let's calculate the maximum value of the capacitor.

in this case, tuning frequency = 1200KHz

$C_{maximum}=\frac{1}{w_{maximum}^2L}=\frac{1}{(2\pi(1200*10^3))^2*200*10^{-6}}=88.04*10^{-12}F$

Hence the maximum  value of the capacitor is 88.04pf

Hence the Range of the values of the capacitor is $88.04pF to 198.1pF$.

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