Question

the variable capacitor in the tuner of an am radio has a capacitance of 2550 pf when the radio is tuned to a station at 550 khz. part a what must the capacitance be for a station near the other end of the dial, 1610 khz? express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Recall the formula for the resonant frequency of an LC circuit

The resonant frequency formula is \( f=\frac{1}{2\pi\sqrt{LC}} \). Since \( L \) (inductance) is constant in this case (the same radio circuit), we can write \( f_1\sqrt{C_1}=f_2\sqrt{C_2} \).

Step2: Rearrange the formula to solve for \( C_2 \)

From \( f_1\sqrt{C_1}=f_2\sqrt{C_2} \), we can get \( C_2 = C_1\times(\frac{f_1}{f_2})^2 \).
Given \( C_1 = 2550\space pF\), \( f_1=550\space kHz\), and \( f_2 = 1610\space kHz\).
Substitute the values into the formula:

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$$ C_2\approx295\space pF$$

Answer:

\(295\space pF\)