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Question

(A) The resonant frequency of an $LC$ circuit is given by $f = \frac{1}{2\pi \sqrt{LC}}$.Rearranging for capacitance,we get $C = \frac{1}{4\pi^2 f^2 L

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$88 \;pF$ to $198 \;pF$

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(A) The resonant frequency of an $LC$ circuit is given by $f = \frac{1}{2\pi \sqrt{LC}}$.Rearranging for capacitance,we get $C = \frac{1}{4\pi^2 f^2 L}$.Given $L = 200 \;\mu H = 200 	imes 10^{-6} \;H$.For $f_1 = 800 \;kHz = 800 	imes 10^3 \;Hz$:$C_1 = \frac{1}{4 	imes (3.14)^2 	imes (800 	imes 10^3)^2 	imes 200 	imes 10^{-6}} \approx 198 \;pF$.For $f_2 = 1200 \;kHz = 1200 	imes 10^3 \;Hz$:$C_2 = \frac{1}{4 	imes (3.14)^2 	imes (1200 	imes 10^3)^2 	imes 200 	imes 10^{-6}} \approx 88 \;pF$.Thus,the range of the variable capacitor is $88 \;pF$ to $198 \;pF$.

$A$ radio can tune over the frequency range of a portion of $MW$ broadcast band: ($800 \;kHz$ to $1200 \;kHz$). 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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