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Chapter 30: Problem 6

The AM radio band covers the frequency range from \(520 \mathrm{kHz}\) to \(1610 \mathrm{kHz}\). Assuming a fixed inductance in a simple LC circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of \(C_{\mathrm{h}} / C_{\mathrm{l}}\) where \(C_{\mathrm{h}}\) is the capacitance for the highest frequency and \(C_{1}\) is the capacitance for the lowest frequency? a) 9.59 b) 0.104 c) 0.568 d) 1.76

Short Answer

Expert verified
Answer: The ratio of capacitance is 0.104.

Step by step solution

01

Find the formula relating capacitance and frequency

In a simple LC circuit, the resonance frequency is given by: $$f = \frac{1}{2\pi \sqrt{LC}}$$ We need to make capacitance, C, the subject of the formula: $$C = \frac{1}{(2\pi f)^2 L}$$ We now have a formula relating capacitance and frequency in an LC circuit.
02

Calculate the capacitance for the highest frequency

We are given the highest frequency, 1610 kHz, and we need to find the corresponding capacitance \(C_\mathrm{h}\). Using the formula from Step 1, we have: $$C_\mathrm{h} = \frac{1}{(2\pi 1610 * 10^3)^2 L}$$
03

Calculate the capacitance for the lowest frequency

Likewise, we are given the lowest frequency, 520 kHz, and need to find the corresponding capacitance \(C_\mathrm{l}\). Using the formula from Step 1, we have: $$C_\mathrm{l} = \frac{1}{(2\pi 520 * 10^3)^2 L}$$
04

Find the ratio of capacitance \(C_\mathrm{h}/C_\mathrm{l}\)

Now, we need to find the ratio of the capacitances. Divide \(C_\mathrm{h}\) by \(C_\mathrm{l}\): $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{\frac{1}{(2\pi 1610 * 10^3)^2 L}}{\frac{1}{(2\pi 520 * 10^3)^2 L}}$$ Simplifying the expression, we get: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2 L}{(2\pi 1610 * 10^3)^2 L}$$ Now, we can cancel out the L from the numerator and the denominator, which leaves us with: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2}{(2\pi 1610 * 10^3)^2}$$
05

Calculate the value of the capacitance ratio

After the simplification, we can now plug the values into the formula to find the final ratio of capacitance: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2}{(2\pi 1610 * 10^3)^2} = 0.104$$ Hence, the correct answer is (b) 0.104.

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Most popular questions from this chapter

In a series RLC circuit, \(V=(12.0 \mathrm{~V})(\sin \omega t), R=10.0 \Omega\) \(L=2.00 \mathrm{H},\) and \(C=10.0 \mu \mathrm{F}\). At resonance, determine the voltage amplitude across the inductor. Is the result reasonable, considering that the voltage supplied to the entire circuit has an amplitude of \(12.0 \mathrm{~V} ?\)

Design an RC band-pass filter that passes a signal with frequency \(5.00 \mathrm{kHz},\) has a ratio \(V_{\text {out }} / V_{\text {in }}=0.500,\) and has an impedance of \(1.00 \mathrm{k} \Omega\) at very high frequencies. a) What components will you use? b) What is the phase of \(V_{\text {out }}\) relative to \(V_{\text {in }}\) at the frequency of \(5.00 \mathrm{kHz} ?\)

A series RLC circuit is in resonance when driven by a sinusoidal voltage at its resonant frequency, \(\omega_{0}=(L C)^{-1 / 2}\) But if the same circuit is driven by a square-wave voltage (which is alternately on and off for equal time intervals), it will exhibit resonance at its resonant frequency and at \(\frac{1}{3}, \frac{1}{5}\), \(\frac{1}{7}, \ldots,\) of this frequency. Explain why.

In Solved Problem 30.1 , the voltage supplied by the source of time-varying emf is \(33.0 \mathrm{~V}\), the voltage across the resistor is \(V_{R}=I R=13.1 \mathrm{~V}\), and the voltage across the inductor is \(V_{L}=I X_{L}=30.3 \mathrm{~V}\). Does this circuit obey Kirchhoff's rules?

An LC circuit consists of a capacitor, \(C=2.50 \mu \mathrm{F},\) and an inductor, \(L=4.0 \mathrm{mH}\). The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, \(R\), is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be \(20 \%\) less than the angular frequency of the oscillations in the LC circuit. a) Determine the resistance of the resistor. b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be \(50 \%\) of the initial amplitude? c) How many complete damped oscillations will have occurred in that time?
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