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Chapter 21: Problem 56

An \(R L C\) series circuit has a resistance of \(R=325 \Omega\) an inductance \(\quad L=0.300 \mathrm{mH}, \quad\) and \(\quad\) a capacitance $C=33.0 \mathrm{nF} .$ (a) What is the resonant frequency? (b) If the capacitor breaks down for peak voltages in excess of \(7.0 \times 10^{2} \mathrm{V},\) what is the maximum source voltage amplitude when the circuit is operated at the resonant frequency?

Short Answer

Expert verified
Answer: The resonant frequency of the RLC circuit is approximately 30.8 kHz and the maximum source voltage amplitude when the circuit is operated at the resonant frequency is approximately 5.41 V.

Step by step solution

01

Find the resonant frequency

To find the resonant frequency, we use the formula for the resonant angular frequency, which is given by \(\omega_0 = \frac{1}{\sqrt{L C}}\). Plug in the given values of L and C to find the resonant angular frequency \(\omega_0\): \(\omega_0 = \frac{1}{\sqrt{(0.300 \times 10^{-3})(33.0 \times 10^{-9})}}\) Now, convert the angular frequency to the resonant frequency \(f_0\): \(f_0 = \frac{\omega_0}{2 \pi}\)
02

Calculate the resonant frequency

Now that we have found the resonant angular frequency, we can calculate the resonant frequency: \(f_0 = \frac{\omega_0}{2 \pi} = \frac{1}{2 \pi\sqrt{(0.300 \times 10^{-3})(33.0 \times 10^{-9})}}\) \(f_0 \approx 30.8 \thinspace kHz\) The resonant frequency of the RLC circuit is approximately 30.8 kHz.
03

Determine the maximum source voltage amplitude

To find the maximum source voltage amplitude when the circuit is operated at the resonant frequency, we can use the given information about the capacitor breaking down and the circuit: \(V_C = QV_s = \frac{I_p}{\omega_0 C}\) We need to find the source voltage amplitude, \(V_s\). Rearranging the equation: \(V_s = \frac{V_C \omega_0 C}{Q}\) We are given \(V_C = 7.0 \times 10^2 \thinspace V\). We need to determine the quality factor of the circuit, Q, which is given by: \(Q = \frac{\omega_0 L}{R}\) Using the given values of L and R, and the calculated value of \(\omega_0\): \(Q = \frac{\omega_0 L}{R} = \frac{1}{R \sqrt{\frac{C}{L}}}\)
04

Calculate the maximum source voltage amplitude

Now, we can plug in the necessary values to find the maximum source voltage amplitude: \(V_s =\frac{(7.0 \times 10^2)}{R \sqrt{\frac{C}{L}}} \times (\frac{1}{\sqrt{L C}}) \times C\) \(V_s =\frac{(7.0 \times 10^2)}{(325)(\sqrt{\frac{33 \times 10^{-9}}{0.300 \times 10^{-3}}})} \times (\frac{1}{\sqrt{(0.300 \times 10^{-3})(33 \times 10^{-9})}}) \times (33 \times 10^{-9})\) \(V_s \approx 5.41 \thinspace V\) The maximum source voltage amplitude when the circuit is operated at the resonant frequency is approximately 5.41 V.

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Most popular questions from this chapter

A parallel plate capacitor has two plates, each of area $3.0 \times 10^{-4} \mathrm{m}^{2},\( separated by \)3.5 \times 10^{-4} \mathrm{m} .$ The space between the plates is filled with a dielectric. When the capacitor is connected to a source of \(120 \mathrm{V}\) rms at \(8.0 \mathrm{kHz}\) an mms current of \(1.5 \times 10^{-4} \mathrm{A}\) is measured. (a) What is the capacitive reactance? (b) What is the dielectric constant of the material between the plates of the capacitor?
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