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Chapter 21: Problem 77

A variable capacitor is connected in series to an inductor with negligible internal resistance and of inductance \(2.4 \times 10^{-4} \mathrm{H} .\) The combination is used as a tuner for a radio. If the lowest frequency to be tuned in is \(0.52 \mathrm{MHz}\). what is the maximum capacitance required?

Short Answer

Expert verified
Answer: The maximum capacitance required for tuning the lowest frequency of 0.52 MHz is approximately 1.95 × 10⁻¹¹ F.

Step by step solution

01

Write the formula for the resonant frequency

The resonant frequency \(f_r\) in a series LC circuit is given by: $$ f_r = \frac{1}{2 \pi \sqrt{LC}}, $$ where L is the inductance, and C is the capacitance.
02

Rearrange the formula to solve for the capacitance

To find the capacitance, we can rearrange the formula as follows: $$ C = \frac{1}{(2 \pi f_r)^2L}. $$
03

Substitute the given values and calculate the maximum capacitance

Now, we can substitute the given values for the lowest frequency and the inductance: The lowest frequency to be tuned in is \(0.52 MHz = 0.52 \times 10^6 Hz\) and the inductor's inductance is \(2.4 \times 10^{-4} H\). So we get: $$ C = \frac{1}{(2 \pi (0.52 \times 10^6))^2 (2.4 \times 10^{-4})} $$ Calculating this value gives: $$ C \approx 1.95 \times 10^{-11} F $$
04

Present the final answer

The maximum capacitance required for tuning the lowest frequency of \(0.52 MHz\) is approximately \(1.95 \times 10^{-11} F\).

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Most popular questions from this chapter

At what frequency is the reactance of a 6.0 - \(\mu\) F capacitor equal to $1.0 \mathrm{k} \Omega ?$
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