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Chapter 30: Problem 74

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.

Short Answer

Expert verified
Answer: To design a series RLC circuit that resonates at 60 Hz with a magnification factor of 20, we need a capacitance of approximately 7.056 x 10^-5 F and a resistance of approximately 37.662 Ohms.

Step by step solution

01

Resonance frequency formula

The resonance frequency, \(f_0\), can be determined using the following formula: $$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$ Here, \(L\) is the inductor value (1 H), \(C\) is the capacitance, and \(f_0\) is the desired resonance frequency (60 Hz).
02

Q-factor formula

The Q-factor, or quality factor, of a series RLC circuit can be determined using the following formula: $$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$ Here, \(R\) is the resistance, and \(Q\) is the Q-factor.
03

Magnification factor formula

The magnification factor, \(M\), for a series RLC circuit is equal to the Q-factor: $$M = Q$$
04

Calculate the capacitance

Using the resonance frequency formula, we can solve for the capacitance: $$C = \frac{1}{(2 \pi f_0)^2 \cdot L}$$ Substitute the given values: $$C = \frac{1}{(2 \pi (60))^2 \cdot 1}$$ $$C \approx 7.056 \times 10^{-5} \mathrm{~F}$$
05

Calculate the Q-factor

Next, we will calculate the Q-factor using the magnification factor formula, since \(M = Q\): $$Q = M = 20$$
06

Calculate the resistance

Now, we can determine the resistance with the Q-factor formula: $$R = \frac{1}{Q} \sqrt{\frac{L}{C}}$$ Substitute the calculated values of \(L\), \(C\), and \(Q\): $$R = \frac{1}{20} \sqrt{\frac{1}{7.056 \times 10^{-5}}}$$ $$R \approx 37.662 \mathrm{~\Omega}$$
07

Design the circuit

Finally, we have determined the necessary values for the series RLC circuit that resonates at 60 Hz with a magnification factor of 20: - Inductor: \(L = 1.00 \mathrm{~H}\) - Capacitor: \(C \approx 7.056 \times 10^{-5} \mathrm{~F}\) - Resistor: \(R \approx 37.662 \mathrm{~\Omega}\) The student should now create a series RLC circuit using these values to achieve the desired resonance frequency and magnification factor.

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Most popular questions from this chapter

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

In the RC high-pass filter shown in the figure, \(R=10.0 \mathrm{k} \Omega\) and \(C=0.0470 \mu \mathrm{F}\) What is the 3.00 -dB frequency of this circuit (where \(\mathrm{dB}\) means basically the same for electric current as it did for sound in Chapter 16 )? That is, at what frequency does the ratio of output voltage to input voltage satisfy \(20 \log \left(V_{\text {out }} / V_{\text {in }}\right)=-3.00 ?\)

Show that the power dissipated in a resistor connected to an AC power source with frequency \(\omega\) oscillates with frequency \(2 \omega\).

A transformer with 400 turns in its primary coil and 20 turns in its secondary coil is designed to deliver an average power of \(1200 .\) W with a maximum voltage of \(60.0 \mathrm{~V}\). What is the maximum current in the primary coil?

A 10.0 - \(\mu\) F capacitor is fully charged by a 12.0 - \(\mathrm{V}\) battery and is then disconnected from the battery and allowed to discharge through a 0.200 - \(\mathrm{H}\) inductor. Find the first three times when the charge on the capacitor is \(80.0-\mu \mathrm{C},\) taking \(t=0\) as the instant when the capacitor is connected to the inductor.
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