Question

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

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

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 \(f_0\): \(f_0 = \frac{\omega_0}{2 \pi}\)

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.

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 \(\omega_0\): \(Q = \frac{\omega_0 L}{R} = \frac{1}{R \sqrt{\frac{C}{L}}}\)

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.