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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

Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a 1.00 - \(\mathrm{H}\) inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of \(60.0 \mathrm{~Hz}\) and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 20.0 times the input voltage or the voltage across the resistor.

What are the maximum values of (a) current and (b) voltage when an incandescent 60 -W light bulb (at \(110 \mathrm{~V})\) is connected to a wall plug labeled \(110 \mathrm{~V} ?\)

A transformer contains a primary coil with 200 turns and a secondary coil with 120 turns. The secondary coil drives a current \(I\) through a \(1.00-\mathrm{k} \Omega\) resistor. If an input voltage \(V_{\mathrm{rms}}=75.0 \mathrm{~V}\) is applied across the primary coil, what is the power dissipated in the resistor?

Design an RC high-pass filter that rejects \(60.0 \mathrm{~Hz}\) line noise from a circuit used in a detector. Your criteria are reduction of the amplitude of the line noise by a factor of 1000 . and total impedance at high frequencies of \(2.00 \mathrm{k} \Omega\) a) What components will you use? b) What is the frequency range of the signals that will be passed with at least \(90.0 \%\) of their amplitude?

Why can't a transformer be used to step up or step down the voltage in a DC circuit?
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