Question

An RLC series circuit has \(L=0.300 \mathrm{H}\) and \(C=6.00 \mu \mathrm{F}\) The source has a peak voltage of \(440 \mathrm{V}\). (a) What is the angular resonant frequency? (b) When the source is set at the resonant frequency, the peak current in the circuit is \(0.560 \mathrm{A}\). What is the resistance in the circuit? (c) What are the peak voltages across the resistor, the inductor, and the capacitor at the resonant frequency?

Short answer

Question: Calculate the angular resonant frequency, the resistance, and the peak voltages across the resistor, inductor, and capacitor in an RLC series circuit with L = 0.300 H, C = 6.00 µF, a source's peak voltage of 440 V, and a peak current of 0.560 A. Answer: The angular resonant frequency is approximately \(2.82 \times 10^3 \tfrac{\text{rad}}{\text{s}}\), the resistance is approximately \(785.71 \Omega\), and the peak voltages across the resistor, inductor, and capacitor are approximately \(439.92 \mathrm{V}\), \(470.33 \mathrm{V}\), and \(33.09 \mathrm{V}\), respectively.

Step-by-step solution

Plug in the values of L and C given in the problem:

L = \(0.300 \mathrm{H}\) and C = \(6.00 \mu \mathrm{F} (6.00 \times 10^{-6} F)\) Now, calculate the angular resonant frequency: \(\omega_0 = \dfrac{1}{\sqrt{(0.300)(6.00 \times 10^{-6})}}\) Step 2: Solve for the angular resonant frequency Solve the expression for the angular resonant frequency: \(\omega_0 = \dfrac{1}{\sqrt{(0.300)(6.00 \times 10^{-6})}} \approx 2.82 \times 10^3 \tfrac{\text{rad}}{\text{s}}\) Step 3: Calculate the resistance in the circuit The source's peak voltage is given as \(440 \mathrm{V}\), and the peak current is given as \(0.560 \mathrm{A}\). At the resonant frequency, the impedance of the circuit (Z) is equal to the resistance (R), so we can use Ohm's law to find the resistance: \(V = IR\)

Plug in the values of V and I given in the problem:

\(V = 440V\) and \(I = 0.560A\) Solve for R: \(R = \dfrac{V}{I} = \dfrac{440}{0.560} \approx 785.71 \Omega\) Step 4: Calculate the peak voltages across the resistor, inductor, and capacitor At the resonant frequency, the peak voltages across the resistor (VR), inductor (VL), and capacitor (VC) are as follows: \(V_R = I \times R\) \(V_L = I \times \omega_0 L\) \(V_C = \dfrac{I}{\omega_0 C}\)

Plug in the values of I, \(\omega_0\), L, and C from the previous steps:

I = \(0.560 \mathrm{A}\), \(\omega_0 = 2.82 \times 10^3 \tfrac{\text{rad}}{\text{s}}\), L = \(0.300 \mathrm{H}\), and C = \(6.00 \times 10^{-6} F\) Calculate the peak voltages: \(V_R = (0.560)(785.71) \approx 439.92 \mathrm{V}\) \(V_L = (0.560)(2.82 \times 10^3)(0.300) \approx 470.33 \mathrm{V}\) \(V_C = \dfrac{0.560}{(2.82 \times 10^3)(6.00 \times 10^{-6})} \approx 33.09 \mathrm{V}\) The peak voltages across the resistor, inductor, and capacitor at the resonant frequency are approximately \(439.92 \mathrm{V}\), \(470.33 \mathrm{V}\), and \(33.09 \mathrm{V}\), respectively.